physics part 1

ALLENÒ CBSE 95 EXERCISE-IV FIVE MARK QUESTIONS : 1. A square loop of sides 5 cm carrying a current of 0.2 A in the clockwise direction is placed at a distance of 10 cm from an infinitely long wire carrying a current of 1 A as shown. Calculate (i) the resultant magnetic force, and (ii) the torque, if any, acting on the loop. Sol. Force on side bc and ad are same in magnitude and opposite in direction. So they cancel each other. ur ur ur ur F2 + F4 = 0 F2 = -F4 Þ Force on side (ab) F1= æ m0I1I2 ö ç 2pd1 ÷´l è ø node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 04_Unit-3.docx F1 = æ 4p ´10-7 ´1´ 0.2 ö ´ 5 www.notesdrive.comç 2p ´ 10 ÷ è ø 0.2A F2 F1 = 2´10-8N (towards left) 1A b c Force on side (cd) F1 F3 d F3 = æ m0I1I2 ö ´ l a 5 cm ç 2pd2 ÷ 10 cm è ø F4 æ 4p ´ 10-7 ´1 ´ 0.2 ö F3 = ç 2p ´ 15 ÷ ´ 5 è ø F3 = 4 ´10-8 N (towards right) 3 Resultant force on loop, F = F1 – F3 (Q F1 & F3 are opp.) = æ 2 - 4 ö ´ 10 -8 çè 3 ø÷ F = 2 ´10-8 N (towards left) 3 (ii) Torque on square loop 'abcd' t = MB sinq {Q q = 0} t=0 2. (i) Depict magnetic field lines due to two straight, long, parallel conductors carrying steady currents I1 and I2 in the same direction. (ii) Write the expression for the magnetic field produced by one of the conductor over the other. Deduce an expression for the force per unit length. E

96 Physics ALLENÒ (iii) Determine the direction of this force. (iv) In figure given below, wire PQ is fixed while the square loop ABCD is free to move under the influence of currents flowing in them. State with reason, in which direction does the loop begin to move or rotate ? PA B I D C Ia dd Q Sol. For part (i), (ii), (iii) refer 03 marks PYQ. 6 www.notesdrive.com (iv) Forces on side AB and CD cancel each other. Now force on AD is attractive due to similar node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 04_Unit-3.docx direction of current and on side BC force is repulsive due to opposite direction of current. As side AD is nearer to wire PQ as compared to BC so force on side AD ( attractive ) is more than the force on side BC ( repulsive), so square loop ABCD does move towards wire PQ. 3. (i) Express Biot-Savart law in the vector form. (Refer to theory topic-1) (ii) Use it to obtain the expression for the magnetic field at an axial point situated at distance d from the centre of a circular coil of radius R carrying current I. (Refer to NCERT Q. no. 10) (iii) Also, find the ratio of the magnitudes of the magnetic field of this coil at the centre and at an axial point for which x = R 3 . Sol. (iii) At this centre of current loop B0 = m0NI 2R At the axial point x = R 3 m 0 NIR 2 m 0 NIR 2 R2 + 3R2 2 ´8R3 ( )B1 = 32 = 2 B0 m0NI B1 \\ = 2R = 8:1 m0NI 2´ 8R 4. (i) With the help of a neat and labelled diagram, explain the principle and working of a moving coil galvanometer. (Refer to theory topic 10) (ii) What is the function of uniform radial field and how it is produced ? (iii) Define current sensitivity of a galvanometer. How is current sensitivity increased ? Sol. (ii) The function of radial field, is to (a) To maximize the deflecting torque acting on the current carrying coil. (b) To increase the strength of magnetic field. E

ALLENÒ CBSE 97 Radial magnetic field is produced by using concave magnetic pole. Also cylindrical soft iron core helps in production of radial magnetic field. (iii) Current sensitivity is the deflection shown by galvanometer for a unit current flow. IS = f or NAB I K Where f is the deflection in the coil. Current sensitivity of galvanometer can be increased by (1) Increasing the number of turns (N) (2) Increasing magnetic induction (B) (3) Increasing area of coil (A) (4) Decreasing the couple per unit twist of the spiral springs. (By using phosphor bronze wire) 5. Amperes law gives a method to cBa®lc.udlrlateofththee magnetic field due to given current distribution. According to it, the resultant magnetic field along a closed boundary circulation Ñò node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 04_Unit-3.docx is equal to µ times the total current crossing the area bounded by the closed curve provided the www.notesdrive.com o electric field inside the loop remains constant. Ampere's law is more useful under certain symmetrical conditions. Consider one such case of a long Straight wire with circular cross- section (radius R) carrying current 1 uniformly distributed across this cross-section. I Line of r force I (i) The magnetic field at a radial distance r from the centre of the wire in the region r > R, is (A) µoI (B) µoI (C) µoIR2 (D) µoIr2 2pr 2pR 2pr 2pR Ans. (A) µoI 2pr (ii) A long straight wire of a circular cross section (radius a) carries a steady current I and the current I is uniformly distributed across this cross-section. Which of the following plots represents the variation of magnitude of magnetic field B with distance r from the centre of the wire? (A) (B) (C) (D) Ans. (A) E

98 Physics ALLENÒ (iii) A long straight wire of radius R carries a steady current I. The current is uniformly distributed across its cross-section. The ratio of magnetic field at R/2 and 2R is (A) 1/2 (B) 2 (C) 1/4 (D) 1 Ans. (D) 1 (iv) A long straight wire of very-very thin radius carries a steady current I. How magnetic field at a distance from this wire changes if value if current in the wire is doubled? (A) Doubled (B) halved (C) three times (D) none of the above Ans. (A) Doubled 6. When the atomic dipoles are aligned partially or fully, there is a net magnetic moment in the direction of the field in any small volume of the material. The actual magnetic field inside material placed in magnetic field is the sum of the applied magnetic field and the magnetic field due to magnetization. This field is called magnetic intensity (II). H= B/µo–M where M is the magnetization of the material µo, is the permittivity of vacuum and B is the total magnetic field. The measure that tells us how a magnetic material responds to an external field is www.notesdrive.com given by a dimensionless quantity is appropriately called the magnetic susceptibility; for a certain node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 04_Unit-3.docx class of magnetic materials, intensity of magnetization is directly proportional to the magnetic intensity. (i) Magnetization of a sample is (A) volume of sample per unit magnetic moment (B) net magnetic moment per unit volume (C) ratio of magnetic moment and pole strength (D) ratio of pole strength to magnetic moment. Ans. (B) (ii) Identify the wrongly matched quantity and unit pair. (A) Pole strength —Am (B) Magnetic susceptibility - dimensionless number (C) Intensity of magnetisation - A/m (D) Magnetic permeability - Henry m Ans. (D) (iii) A bar magnet has length 3 cm, cross-sectional area 2 cm2 and magnetic moment 3 A m2. The intensity of magnetisation of bar magnet is (A) 2 × l05A/m (B) 3 ×105 A/m (C) 4 × 105 A/m (D) 5 × 105A/m Ans. (D) (iv) A solenoid has core of a material with relative permeability 500 and its windings cany a current of 1 A. The number of turns of the solenoid is 500 per metre. The magnetization of the material is nearly :- (A) 2.5 × 103A/m (B) 2.5 × 105 A/m (C) 2 × 103 A/m (D) 2 ×105 A/m Ans. (B) (v) The relative permeability of iron is 6000. Its magnetic susceptibility is (A) 5999 (B) 6001 (C) 6000 × 10–7 (D) 6000 × 10–7 Ans. (A) E

ALLENÒ CBSE 99 EXERCISE-V (RACE) CHAPTER-4 : MOVING CHARGES & MAGNETISM For question numbers 1 to 5 two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below. (a) Both A and R are true and R is the correct explanation of A (b) Both A and R are true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false and R is also false 1. Assertion : Magnetic moment of toroid is zero. Reason : Magnetic field outside the volume of current carrying toroid is zero. node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 04_Unit-3.docx (i) a (ii) b (iii) c (iv) d www.notesdrive.com 2. Assertion : Two charge particle at rest experience only electrostatic force. Reason : Charges at rest can only produces electrostatic field. (i) a (ii) b (iii) c (iv) d 3. Assertion : A moving charged particle gets energy from magnetic field. Reason : Magnetic force works on moving charged particle. (i) a (ii) b (iii) c (iv) d 4. Assertion : The coil is wound over the metallic frame in moving coil galvanometer. Reason : The metallic frame help in making steady deflection without any oscillation. (i) a (ii) b (iii) c (iv) d 5. Assertion : When magnet is brought near iron nails, only translatory force act on it. Reason : The field due a magnet is generally uniform. (i) a (ii) b (iii) c (iv) d 6. An electron does not suffer any deflection while passing through a region of unifrom magnetic field, what is the direction of the magnetic field ? 7. Write the relation for the force acting on a charged particle q moving with velocity in the presence of magnetic field Br . 8. Write the underlying principle of a moving coil galvanometer. 9. What can be the cause of helical motion of a charged particle? 10. State Ampere's circuital law. 11. Why does a moving charge experience a force when placed in a magnetic field? 12. State two properties of the material of the wire used for suspension of the coil in MCG. 13. What is the force acting an electron when moving (i) parallel and (ii) anti-parallel to the magnetic field? 14. An electron beam projected along x-axis experiences a force due to a magnetic field along Y- axis. What is the direction of the magnetic field? 15. How will the magnetic field intensity at the centre of a circular coil carrying current change if the current through the coil is double and the radius of the coil is halved? E

100 Physics ALLENÒ 1. (ii) RACE CHAPTER-4 (SOLUTIONS) 2. (i) 3. (iv) 4. (i) 5. (iv) 6. Fr = qvB sin q q = 0° or 180° 7. Fr = q (vr ´Br ) www.notesdrive.com 8. Moving coil galvanometer works on the principle that when a current carrying coil placed in a magnetic field, it experiences at torque so, it get deflected and by measuring deflection we can node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 04_Unit-3.docx measure current. 9. When a charged paraticle move in uniform external magnetic field, with velocity not perpendicular or parallel to the magnetic field, then charge particle experiences a force also along with torque and perform helical motion. 10. The line interal of magnetic field over a closed loop is m0 times the total current threading through the loop. Ñò Br . d r = m 0 ( SI ) l 11. A moving charge produces a magnetic field. This magnetic field interacts with another magnetic field of a magnet and hence, it experiences force. 12. (i) High tensile strength (ii) Small torsional constant or small restoring torque per unit twist. 13. Zero. 14. Along the +Z axis (Br = Bkˆ); According to Fleming's left hand rule. 15. B = m0 NI ; Magnetic field becomes 4 times. 2r E

ALLENÒ CBSE 101 EXERCISE-V (RACE) CHAPTER-5 : MAGNETISM AND MATTER For question numbers 1 to 5 two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below. (a) Both A and R are true and R is the correct explanation of A (b) Both A and R are true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false and R is also false 1. Assertion : Magnetic moment of helium atom is zero. Reason : All the electron are paired in helium atom orbitals. (i) a (ii) b (iii) c (iv) d 2. Assertion : An atom behaves as magnetic dipole. node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 04_Unit-3.docx Reason : It is because, an atom contains equal positive & negative charges. www.notesdrive.com (i) a (ii) b (iii) c (iv) d 3. Assertion : When radius of circular loop carrying current is doubled, its magnetic moment becomes four times. Reason : Magnetic moment depends on area of the loop. (i) a (ii) b (iii) c (iv) d 4. How is a toroid different from a solenoid? 5. Is the steady current the only source of magnetic field ? 6. Which orientation of a magnetic dipole in a uniform magnetic field will correspond to its stable equilibrium? 7. Must every magnetic configuration have a north pole and a south pole? What about the field due to a toroid? 8. Does a bar magnet exert a toque on itself due to its own field? Does on element of a current- carrying wire exert a force on another element of the same wire? 9. How does the (i) pole strength, and (ii) magnetic moment of each part of a bar magnet change if it is cut into two equal pieces transverse to its length? 10. What happens if a bar magnet is cut into two pieces: (i) transverse to its length (ii) along its length RACE CHAPTER-5 (SOLUTIONS) 1. (i) 2. (iii) 3. (ii) 4. A solenoid has N-S poles whereas toroid doesn't have separate poles. 5. No, magnet and unsteady current will also be source of magnetic field. 6. In stable equilibrium, the dipole moment vector and the magnetic field vector are in same direction. 7. Not necessarily. True only if the source of the field has a net non-zero magnetic moment. This is not so for a toroid or even for a straight infinite conductor. 8. No. There is no force or torque on a element due to the field produced by that element itself. But there is a force (or torque) on an element of the same wire. 9. (i) The pole strength does not change. (ii) The magnetic moment reduces to half. 10. In either case, one gets two magnets, each with a north and south pole. E

102 Physics ALLENÒ EXERCISE - VI (MOCK TEST) CHAPTER-4 : MOVING CHARGES & MAGNETISM 1. How will the magnetic field intensity at the centre of a circular coil carrying current change if the current through the coil is double and the radius of the coil is halved? [1] 2. A proton and an electron travelling along parallel paths enter a region of uniform magnetic field, acting perpendicular to their paths. Which of them will move in a circular path with higher frequency? [1] 3. State two properties of the material of the wire used for suspension of the coil in MCG. [1] 4. What is radial magnetic field? How it is obtained in moving coil galvanometer? [2] 5. Two wires of equal lengths are bent in the form of two loops. One of the loop is square shapedwww.notesdrive.com whereas the other loop is circular .These are suspended in a uniform magnetic field and the same current is passed through them. Which loop will experience greater torque? Give reasons? [2] node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 04_Unit-3.docx 6. A steady current flows in the network shown in the figure. What will be the magnetic field at the centre of the network? [2] Q PR S 7. State two properties of the material of the wire used for suspension of the coil in a moving coil galvanometer? 8. A circular coil is placed in uniform magnetic field of strength 0.10T normal to the plane of coil. If current in the coil is 5.0A. Find. (a) Total torque on the coil (b) Total force on the coil (c) Average force on each electron due to magnetic field (The coil is made of copper wire of cross- sectional area 10–5 m2 and free electron density in copper is 1029 m–3) 9. A particle of mass m and charge q moving with a uniform speed u normal to a uniform magnetic field B describes a circular path of radius & Derive expressions for (1) Radius of the circular path (2) time period of revolution (3) Kinetic energy of the particle? 10. Write the formula of Biot-Savart's law in vector form. Obtain an expression of magnetic field on the axis of a current carrying circular loop. Draw necessary diagram. 11. Read the case study carefully and answer the questions that follow : E

ALLENÒ CBSE 103 11. Read the case study carefully and answer the questions that follow : A galvanometer is a device used to detect current in an Ammeter electric circuit. It cannot as such be used as an ammeter to S measure current in a given circuit. This is because a (I–Ig) (I–Ig) galvanometer is a very sensitive device, It gives a full- scale defleclion for a current of the order of µA. I A Ig G Ig B I Moreover, for measuring currents, the galvanometer has to be connected in series, and it has a large resistance, this will change the value of current in the circuit. To overcome these difficulties, we connect a small resistance Rs, called shunt resistance, in parallel with the galvanometer coil, so that most of the current passes through the shunt. Now to use galvanometer as a voltmeter, it has to he connected in parallel with the circuit element across which we need to measure p.d. Moreover, it must draw a very small current, otherwise it will appreciably change the voltage which we are measuring. To ensure this a large resistance R is connected in series with the galvanometer. node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 04_Unit-3.docx Voltmete www.notesdrive.com G R A Ig B V Based on the information given above, answer the following questions: (i) A sensitive galvanometer like a moving coil galvanometer can be converted into an ammeter or a voltmeter by connecting a proper resistance to it. Which of the following statements is true:- (A) a voltmeter is connected in parallel and current through it is negligible (B) an ammeter is connected in parallel and potential difference across it is small (C) a voltmeter is connected in series and potential difference across it is small (D) an ammeter is connected in series in a circuit and the current through it is negligible. (ii) By mistake a voltmeter is connected in series and an ammeter is connected in parallel with a resistance in an electrical circuit. What will happen to the instruments? (A) Voltmeter is damaged (B) Ammeter is damaged (C) Both are damaged (D) None is damaged. (iii) A galvanometer coil has a resistance of 15 W and gives full scale deflection for a current of 4 mA. To convert it to an ammeter of range 0 to 6 A. (A) 10 m W resistance is be connected in parallel to the galvanometer. (B) 10 m W resistance is to be connected in series with the galvanometer. (C) 0.1 W resistance is to be connected in parallel to the galvanometer. (D) 0.1 W resistance is to be connected in series with the galvanometer. (iv) Two identical galvanometers are converted into an ammeter and a milli ammeter. Resistance of the shunt of milliammeter through which the current passes will be (A) more (B) equal (C) less (D) zero E

104 Physics ALLENÒ EXERCISE - VI (MOCK TEST) CHAPTER-5 : MAGNETISM AND MATTER 1. What happens if a bar magnet is cut into two pieces: [1] (i) transverse to its length, (ii) along its length? [1] 2. Does the magnetic moment for solenoid and corresponding bar magnet is equal? 3. An iron ring of relative permeability µr has windings of insulated copper wire of n turns per metre. When the current in the windings is I, find the expression for the magnetic field in the ring. [2] 4. A short bar magnet placed with it's axis at 30° to a uniform magnetic field of 0.2T experiences a www.notesdrive.com torque of 0.060 N-m. [2] node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 04_Unit-3.docx (i) Calculate magnetic moment of the magnet and (ii) find out what orientation of the magnet corresponds to it's stable equilibrium in the magnetic field. 5. Prove that the magnetic moment of the electron revolving around a nucleus in an orbit of radius r with orbital speed v is equal to evr/2. Hence using Bohr's postulate of quantization of angular momentum, deduce the expression for the magnetic moment of hydrogen atom in the ground state. [3] 6. A short bar magnet of magnetic moment 0.32 J/T is placed in uniform field of 0.15 T. If the bar is free to rotate in plane of field then which orientation would correspond to its (i) stable and (ii) unstable equilibrium? What is potential energy of magnet in each case? [3] 7. A solenoid has a core of a material with relative permeability 400. The windings of the solenoid are insulated from the core and carry a current of 2A. If the number of turns is 1000 per metre, calculate (a) H, (b) Intensity of magnetisation (I), (c) B and (d) the magnetising current Im. [5] E

ALLENÒ CBSE 105 UNIT-IV : ELECTROMAGNETIC INDUCTION & ALTERNATING CURRENT CHAPTER-6 : ELECTROMAGNETIC INDUCTION 1. ELECTROMAGNETIC INDUCTION (EMI) The phenomena in which emf is induced by varying magnetic field is called EMI. 2. MAGNETIC FLUX (f) f = NBr.Ar = NBA cos q SI unit : weber (Wb) or T-m2 magnetic flux is scalar quantity 3. FARADAY'S LAW It states that the magnitude of induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit. node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docx Induced emf = - N df www.notesdrive.com dt 4. Lenz's LAW It states that the direction of current induced in the loop is such that it opposes the cause due to which it get induced. It is in accordance with law of conservation of energy. Induced emf = –N df here negative sign is due to Lenz law dt Charge flown due to induced emf, Q = Df R 5. MOTIONAL EMF When a rod moves perpendicular to its length and perpendicular to magnetic field then emf induced in the rod is called motional emf. r Br ) (vr ^ l ^ BÄ l V Motional induced emf e = BlV When a conducting disc or conducting rod is rotated about its axis ^ to magnetic field then emf induced between its centre and periphery is given by emf = Bwl2 2 BÄ BÄ w ll E

106 Physics ALLENÒ 6. EDDY CURRENTS When any metal (conductor of any shape) is subjected to changing magnetic flux, induced current produced in the form of small loops (eddies) are called eddy currents. Its direction is given by Lenz law. Advantage / Application (i) Magnetic braking system (ii) Electromagnetic damping (iii) Induction motor (iv) Induction furnace Disadvantage Eddy current heat up the core of conductor and thus dissipated the energy. 7. INDUCTANCE It is the ratio of flux linkage to current. It is equal to Nf / I. It is scalar qty. SI unit Henry (H) it depends on geometry of the conductor & intrinsic magnetic properties. 7.1 Self induction and Self Inductance : Phenomena of induced emf due to change in its own current is called self induction. www.notesdrive.com N f = Li, emf = - Ldi node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docxdt Self inductance (L) : Self inductance is numerically equal to the magnetic flux linked with coil when a unit current flows through it. Self inductance (L) for a solenoid : L = m0N2A = m0n2 Al l N = total number of turns; n = number of turns/length Combination of inductors Series L = L1+ L2 Parallel 1 = 1 1 L L1 + L2 7.2 Mutual induction & mutual inductance : Phenomena of inducing emf in a coil due to change in current in another coil is known as mutual induction. N f = Mi emf = -M di dt Mutual inductance (M) : Mutual inductance is numerically equal to the magnetic flux linked with one coil when a unit current flows through the other coil. Mutual inductance between 2 solenoids M = m0N1N2A l Mutual induction of the pair of coils, solenoids depends on their separation as well as their relative orientation. E

ALLENÒ CBSE 107 Relation between self inductance and mutal inductance. L2 M = k L1L2 M = L1L2 L1 k = coupling factor k = 1 for ideal coil air gap soft iron 0<k<1 k=1 8. ENERGY STORED IN INDUCTOR U = 1 Li2 2 node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docxEnergydensity=B2 2µ0 www.notesdrive.com 9. AC GENERATOR AC generator works on the principle of EMI Coil Axle Slip Alternating emf rings Carbon brushes e = NBAwsin wt e = e0 sin wt 10. IMPORTANT vPOINTS (i) N S Loop will repel vthe magnet (ii) N S Loop will attract the magnet (iii) An emf is induced in a closed loop when magnetic flux tihsevlainrieedin. tTeghrealinÑòduErc.derld electric field is non conservative field because for induced electric field, around a closed path is non-zero. (iv) Acceleration of a magnet falling through a long solenoid decreases because the induced current produced in a circuit always flows in such a direction that it opposes the change or the cause which produced it. E

108 Physics ALLENÒ CHAPTER-7 : ALTERNATING CURRENT 1. Nature Wave–form (voltage, current) Sinusoidal 0 + – T V = V0 sin wt p 2p I = I0 sin wt 2. ROOT MEAN SQUARE VALUE OF AC CURRENT (RMS)/ EFFECTIVE CURRENT (Irms) Irms = I0 = 0.707 I0 , I0 = peak value of current www.notesdrive.com2 3. AVERAGE OR MEAN VALUE node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docx IAvg = 2I0 = 0.637 I0 p 4. PHASORS AND PHASOR DIAGRAM A phasor is a vector which rotates about the origin with angular speed w. Phasor diagram shows the phase relationship between voltage & current. Vm sinwt1 Vm Ii m V I 0 wt1 2p wt Im sinwt1 Phasor Diagram 5. AC VOLTAGE APPLIED TO A PURE RESISTANCE (R) R V = V0 sin wt i = V0 sin wt R Resistance = R i VR, R VR = iR voltage and current are in same phase. E

ALLENÒ CBSE 109 6. AC VOLTAGE APPLIED TO A PURE INDUCTOR (L) L V = V0 sin wt i = V0 (- cos wt) wL Reactance XL = wL XL,VL i VL = iXL voltage leads the current by p/2 node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docx 7. AC VOLTAGE APPLIED TO A PURE CAPACITOR (C) www.notesdrive.com C V = V0 sin wt i = V0 cos wt (1/ wC) Reactance XC = 1 wC i XC, VC VC = iXC current leads voltage by p/2 8. AC THROUGH LCR CIRCUIT VL VR i R LC XL–XC v=V0sinwt VC Z (a) If VL > VC Þ XL > XC then VL–VC V= VR2+(VL–VC)2 f f VR R E

110 Physics fR ALLENÒ E (b) If VC > VL Þ XC > XL then f VR Z VC–VL V= VR2+(VL–VC)2 XC–XL (c) If VC = VL Þ XC = XL then V = VR , Z = R and f = 0 9. IMPEDENCE Impedence = Z = R2 + (X L - X C )2 10. POWER IN AC CIRCUITwww.notesdrive.com V = V0sin wt node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docxfi i = I0 sin(wt + f) V Power = Vrms irms cos f = i2 R Wattfull current = irms cosf Wattless current = irms sinf Wattless power = vrms irms sinf Where cosf = Power factor 11. RESONANCE IN SERIES LCR CIRCUIT At resonance • XL = XC or VL = VC • Z = R = min Þ i= V = max R • Power factor æ cos f = R 1øö÷ çè Z= • Angle (or phase difference) between v and i = 0° • VR = VSource Resonating frequency w0 = 1 LC i R1 > R2 R2 Z xC>xL xL>xC R1 xL=xC w w0 w w0 Sharpness of resonance / quality factor = XL = 1 L = f0 R R C Band width

ALLENÒ CBSE 111 12. TRANSFORMER Laminated Secondary It transfers electric power. Soft iron core winding Works only for AC Principle : Mutual induction Primary For ideal transformer winding (1) Power loss = 0 Þ efficiency = 100% AC mains (2) Flux loss = 0 node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docx But practicaly Pout < Pin \\ efficiency < 100% www.notesdrive.com LoadVS=NS=ip= turnratioh(efficiency)=is2R´ 100 VP NP is VP ip Types of transformer Step up Step down VS > VP VP > VS NS > NP NP > NS iP > iS iS > iP Transmissions are done at high voltage and low current by using step up transformer. Losses in tranformers Coil (Cu losses) Core (Iron losses) Heat loss Flux loss Hysteresis Eddy currents Can be minimize Can zero by Can be controlled by using Can be controlled by using thick wires making coupling factor1 substance of high mr (soft iron) by laminating the core 13. IMPORTANT POINTS • An alternating current of frequency 50 Hz becomes zero, 100 times in one second because alternating current changes direction and becomes zero twice in a cycle. • An alternating current cannot be used to conduct electrolysis because the ions due to their inertia, cannot follow the changing electric field. • Average value of AC is always defined over half cycle because average value of AC over a complete cycle is always zero. • AC current flows through the surface of wire instead of flowing through total volume of wire. This is known as skin effect. Note : Important datas for LCR series circuit. • Irms = Vrms = (Vrms )L = (Vrms )C = (Vrms )R R LC Z XL XC R • Vrms = (Vrms )2R + [(Vrms )L - ( Vrms )C ]2 • tanf = | VL - VC | = |X L - XC| V=V0sinwt VR R • <P> = Vrms Irms cosf = RVr2ms Z2 E

112 Physics ALLENÒ NCERT IMPORTANT QUESTIONS CHAPTER-6 : ELECTROMAGNETIC INDUCTION 1. A circular coil of radius 10 cm, 500 turns and resistance 2 W is placed with its plane perpendicular to the horizontal component of the earth’s magnetic field. It is rotated about its vertical diameter through 180° in 0.25 s. Estimate the magnitudes of the emf and current induced in the coil. Horizontal component of the earth’s magnetic field at the place is 3.0 × 10–5 T. Sol. Initial flux through the coil, fB(initial) = BA cos q = 3.0 × 10–5 × (p ×10–2) × cos 0º = 3p × 10–7 Wb Final flux after the rotation, www.notesdrive.com fB(final) = BA cos1800 = 3.0 × 10–5 × (p ×10–2) × cos 180° node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docx = –3p × 10–7 Wb Therefore, estimated value of the induced emf is, e = N Df Dt e = 500 ´(6p´10-7 ) = 3.8 × 10–3 V 0.25 Induced current I = e = 1.9´10-3 A R 2. A long solenoid with 15 turns per cm has a small loop of area 2.0 cm2 placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing? Sol. Here, number of turns per unit length, n= N/l = 15 turns/cm=1500 turns/m A = 2 cm2 = 2 × 10–4 m2 dI/dt = (4 – 2)/0.1 or dI/dt = 20 As–1 e = df = d (BA) êéëQ B = m0 NI ù dt dt l ûú e =Addt æ m 0 NI ö =Am0 æ N ö dI èç l ø÷ èç l ø÷ dt e = (2 × 10–4) × 4p × 10–7 × 1500 × 20 V e = 7.5 × 10–6 V E

ALLENÒ CBSE 113 3. A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil? Sol. Magnetic flux f1= MI1 As I1 = 0 \\ f1= 0 f2 = MI2 = 1.5 × 20 = 30 Change in flux df = f2 – f1 = 30 – 0 = 30 weber. 4. Predict the direction of induced current in the situations described by the following figs. (a) and (b). node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docxSol. (a) Using lenz’s rule ,the direction of induced current is along qrpq www.notesdrive.com(b) Using lenz’s rule ,the direction of induced current is along prqp 5. Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit. Sol. Change in current dI= 0 – (5) = –5A Change in time dt = 0.1 s average emf e = 200 V Þ e = -L dI dt 200 = L æ 5ö Þ L = 20 = 4H çè 0.1 ÷ø 5 6. Use Lenz’s law to determine the direction of induced current in the situations described by fig. : (a) A wire of irregular shape turning into a circular shape; (b) A circular loop being deformed into a narrow straight wire. E

114 Physics ALLENÒ Sol. According to Lenz’s law, the direction of the induced emf is such that it tends to produce a current that opposes the change in the magnetic flux that produced it. (a) When the shape of the wire changes, the flux piercing through the unit surface area increases. As a result, the induced current produces an opposing flux. Hence, the induced current flows along adcb. (b) When the shape of a circular loop is deformed into a narrow straight wire, the flux piercing the surface decreases. Hence, the induced current flows along a’d’c’b’. 7. A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s–1 in a uniform horizontal magnetic field of magnitude 3.0 × 10–2 T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 10 W, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from? Sol. Area of the coil, A = pr2 = p( 0.08)2 number of turns N = 20 turns magnetic field B = 3.0 × 10–2 T www.notesdrive.com angular speed w = 50 rad s–1 node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docx Total resistance produced by the loop,R = 10 W Maximum emf induced is given as: I = e = 0.603 = 0.0603A R 10 e = NBAw = 20 × 3.0 × 10–2 × p( 0.08)2 × 50 = 0.603 V The maximum emf induced in the coil is 0.603 V. Over a full cycle, the average emf induced in the coil is zero. Maximum current is given as: Average power because of the Joule heating: P = eI = 0.603´0.0603 = 0.018W 2 2 The torque produced by the current induced in the coil opposes the normal rotation of the coil. To keep the rotation of the coil continuous, we must find a source of torque which opposes the torque by the emf, so here the rotor works as an external agent. Hence, dissipated power comes from the external rotor. 8. Define mutual inductance between a pair of coils. Derive an expression for the mutual inductance of two long coaxial solenoids of same length wound one over the other. Sol. Mutal inductance : It is numerically equal to the primary coil magnetic flux linked with one coil (secondary coil) when unit current flows through the other coil. Derivation :- II (I2, B2, N2, A2) ForI coil, f1= M12 I2 ...(1) Coil where f1 is the flux through I coil. Similarly, f1 = (B2.A1)N1 CoI il (I1, B1, N1, A1) = æ µ0 N2I2 ö A1N1 èç l ÷ø = µ0N1N2A1I2 ...(2) l E

ALLENÒ CBSE 115 From (1) & (2) M12I2 = µ0 N1N 2 A1I2 l Þ M12 = µ0 N1N2A1 l Now, f2 = M21I1 ...(3) where f2 is the flux through 2nd coil due to 1st coil. Also, f2 = B1 A1 N2 cosq = µ0N1I1A1N2 ...(4) (Q q = 0º) l node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docxFrom (3) & (4) www.notesdrive.comM21I1 =µ0N1N2 A1I1 l M21 = µ0 N1N 2 A1 l \\ M12 = M21 9. Define self-inductance of a coil. Obtain the expression for the energy stored in an inductor L connected across a source of emf. Sol. Self inductance (L):- It is numerically equal to the induced emf per unit rate of change of current through a given coil. i.e. L= e - dI dt Derivation :- L We know, dw = Vdq Also, dw = –edq (can be written) Qe = LdI dt So, dw = edq +– E dw = LdI .dq dt or dw = Ld I. dq dt wI (taking limits on both sides) òdw = Lò IdI 00 U= w = LI2 = Energy in inductor 'L' 2 E

116 Physics ALLENÒ 10 . Derive the expression for the magnetic energy stored in an inductor when a current I develops in it. Hence, obtain the expression for the magnetic energy density. Sol. Energy stored in an inductor: Work has to be done by battery against the opposing induced emf in establishing a current in an inductor. The energy supplied by the battery is stored in the inductor. L Let current flowing through the circuit at any instant (t) be i. Rate of change of current at that time = di dt magnitude of induced emf in the inductor, ii e = L di dt Work done by the battery in time dt, +– dW = Vdq = edq V dW = eidt www.notesdrive.com Þ dW = æ L di ö idt node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docxçèdt÷ø dW = Lidi Total work done by the battery in increasing current from 0 to I òW = L i di Þ W = 1 LI2 2 This work done is stored in an inductor in the form of magnetic energy. UB = 1 LI2 2 Energy density (uB) ® Energy stored per unit volume in the magnetic field of an inductor is know as energy density. uB = Energy stored in an inductor Volume 1 LI2 1 ´ æ m0 N2 A ö ´ æ Bl ö2 ìïïQ B = m0NI , 2 ç l ÷ ç m0N ÷ l uB = 2 Þ= è ø è ø A´l í A´l ï L = m0N2A ïî l uB = B2 2m0 11. A conducting rod of length 'l' is moving with constant linear speed 'v' in a uniform magnetic 'B'. This arrangement is mutually perpendicular. Obtain the expression of motional electromotive force. Sol. Expression of motional electromotive Force :- Let according to diagram, a conducting rod PQ of length l is moving with velocity (v) in plane of paper. A magnetic field B is perpendicular into the plane of paper. Free electrons present in conductor are also moving perpendicular to the magnetic field. Lorentz force acting on free electrons is given by - Fm = evB sin 90° Fm = evB E

ALLENÒ CBSE 117 The direction of this force is given by Fleming left hand rule and acts towards end 'Q'. Due to this force negatively charged free electrons move towards end 'Q' and leaving positive charged ions at 'P'. Resultantly the end 'P' becomes positively charged and the end 'Q' becomes negatively charged, which develops a potential difference (e) across the ends of the conductor. So, electric field gets induced in the conductor æ E = e ö . çè l ÷ø Now, this electric field exerts electric force on electrons P B i.e. Fe = eE (towards end P) Fe In equilibrium state Fe = Fm lv eE = evB Fm ---- e´e =evB Q l Motional emf E = v B l node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docx www.notesdrive.com 12. Two concentric circular coils of radii r and R are placed coaxially with centres coinciding. If R >> r then calculate the mutual inductance between the coils. Sol. Let current (I1) is flowing in first outer coil. The magnetic field at centre of coil B1 =m20RI1 Flux passing through second inner coil f2 = B1A2 f2 = m 0 I1 ´ pr 2 2R From definition of mutual induction f2 = M21I1 I1 r M21 = f2 = m0pr2 I1 2R M21 = m0pr2 2R Similarly, M12 = m 0 pr 2 R 2R R >>> r M12 = M21 = M M = m0pr2 2R E

118 Physics ALLENÒ NCERT IMPORTANT QUESTIONS CHAPTER-7 : ALTERNATING CURRENT 1. A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series L-C-R circuit in which R = 3 W, L = 25.48 mH, and C = 796 mF. Find (a) the impedance of the circuit; (b) the phase difference between the voltage across the source and the current; (c) the power dissipated in the circuit. Sol.(a) To find the impedance of the circuit, we first calculate XL and XC. XL =wL =2 pnL = 2 × 3.14 × 50 × 25.48 × 10–3 = 8 W XC = 1 = 1 = 1 = 4W wC 2pnC 2 ´3.14 ´50 ´ 796´10-6 Z = R2 + (X L - XC )2 = 32 + (8 - 4)2 = 5 ohm (b) Phase difference, www.notesdrive.com f = tan -1 æ XC - XL ö node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docxçèRø÷ f = tan -1 æ 4 - 8 ö = tan-1 æ -4 ö = -53° èç 3 ÷ø çè 3 ÷ø Since f is negative, the current in the circuit lags the voltage across the source. (c) The power dissipated in the circuit is P = I2R Now, I= im = 1 æ 283 ö = 40A 2 2 çè 5 ÷ø Therefore, P = (40)2 × 3= 4800W 2. Obtain the resonant frequency wr of a series L-C-R circuit with L = 2.0H, C = 32 mF and R = 10 W. What is the Q-value of this circuit? Sol. Using, wr = 1 ,we get LC wr = 1 = 8 1 -3 = 125s-1 2´ 32´ 10-6 ´10 Then , Q= XL = wr L = 125´ 2 = 25 R R 10 3. An inductor L,a capacitor 20 μF, a resistance 10W.are connected in series with an ac source of frequency 50 Hz. If the current is in phase with voltage, calculate the Inductance L. Sol. Current and voltage are in phase. tanf = æ XL - XC ö = 0 çè R ø÷ Þ XL = XC Þ L = 1 = 1 w2C 4p2 f 2 C L= 1 = 0.5 H (2 ´ 3.14 ´ 50)2 ´ 20 ´10-6 E

ALLENÒ CBSE 119 4. A series LCR circuit with R = 20 W, L = 1.5 H and C = 35 mF is connected to a variable- frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle? Sol. At natural frequency XL = XC \\ Z = R Using , P = V2 , we get R P= 200 ´ 200 = 2000 W 20 5. A 60 mF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit. Sol. Here , XC = 1 = 1 = 2p ´ 60 1 ´10-6 wC 2pnC ´ 60 node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docx \\ I rms = Vrms = æ 110 ö www.notesdrive.comXC 1 çè 2p´ 60 ´ 60 ´10-6 ø÷ = 110 × 2p × 60 × 60 × 10–6 = 2.49 A 6. Describe any two energy losses occuring in transformers. How these can be minimized ? Why electric power is transmitted at high voltage up to large distances ? Sol. (1) Hysteresis loss % Transformer cores are magnetised and demagnetised repeatedly, so lot of electrical energy is wasted in the form of heat in this process. To reduce this loss cores are made of soft iron, which has small hysteresis loop area. (2) Eddy current loss % The alternating magnetic flux induces eddy currents in the iron core causes heating. The effect is reduced by laminating the core. To reduce electrical energy losses in the form of heat energy, power is transmitted at high voltage upto large distances. 7. Write one merit and one de-merit of alternating current in comparison to direct current. Obtain expression for following in a pure inductive alternating current circuit- (i) Instantaneous value of current (ii) Reactance of circuit (iii) Peak value of current Draw curve for power in pure inductive circuit. Sol. Advantage of ac over dc :- With the help of a transformer, ac at any desired voltage can be obtained. That is why power wastage in ac transmission is very low. Disadvantage of ac over dc :- The ac current is more dangerous than the dc because its peak value is 2 times of rms value. E

120 Physics ALLENÒ Pure inductive ac circuit :- VL (i) Instantaneous value of current : L Instantaneous value of applied voltage V = V0 sin wt ....(1) ~ V = V0 sin wt If instantaneous current in inductor is i, then induced emf VL = -L di dt {–ve sign indicates that induced emf oppose the applied voltage Applying KVL, V + VL = 0 V0 sin wtwww.notesdrive.com-Ldi=0 dt node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docx di = V0 sin wt dt L by integration i= V0 ò sin wt dt L i = V0 ëéê- cos wt ù L w ûú i = V0 sin( wt - p / 2) ïì V0 = i0 (peak value) wL í wL ïî i = i0 sin(wt - p / 2) (ii) Reactance of a circuit : i0 = V0 Þ wL = V0 Þ XL = wL wL i0 where XL is reactance of a circuit (iii) Peak value of current : i0 = V0 XL Power curve :- Power ÅÅ 0 T/4 T/2 3T/4 T t E

ALLENÒ CBSE 121 8. (i) Draw a labelled diagram of a step-down transformer. State the principle of its working (ii) Express the turn ratio in terms of voltages. (iii) Find the ratio of primary and secondary currents in terms of turn ratio in an ideal transformer. (iv) How much current is drawn by the primary of a transformer connected to 220 V supply when it delivers power to a 110 V – 550 W refrigerator ? Sol. (i) Np Ns Secondary Primary node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docxNs < Np www.notesdrive.comPrinciple: Transformer works on principle of mutual induction , in which an EMF is induced in the secondary coil by the change in magnetic flux in primary coil. Working : When alternating current source is connected to the ends of primary coil, the current changes continuously in the primary coil, due to which magnetic flux linked with the secondary coil changes continuously. Therefore, the alternating emf of same frequency is developed across the secondary terminals. For part (ii) and (iii) refer three marks previous year question No. 4 (iv) Given VP = 220 V VS = 110 V P = 550 W , Ip = to find IP = Power = P Primary Voltage VP IP = 550 = 2.5 A 220 9. Define the following : (i) Root mean square value of A.C. (ii) Quality factor in electrical resonance. Sol. (i) Root mean square value of alternating current : It is equal to that value of steady current which when passed through same resistance for same time then same amount of heat get produced. (ii) Quality factor in electrical resonance : It is that factor which represents the sharpness of resonance of series LCR circuit. It is given by - Q = 1 L R C E

122 Physics ALLENÒ PREVIOUS YEARS QUESTION CHAPTER - 6 : ELECTROMAGNETIC INDUCTION CHAPTER - 7 : ALTERNATING CURRENT EXERCISE-I ONE MARK QUESTIONS : 1. Give one use of eddy currents. Sol. Eddy currents are used in induction furnace. Strong eddy currents produces desired heating. 2. Explain why current flows through an ideal capacitor when it is connected to an a.c. source but not when it is connected to a d.c. source in a steady state. Sol. When AC is connected to capacitor, due to continuous change of polarity of the applied voltage there will be continuous change of polarity of capacitor plates. This causes the charge to flow across the capacitor. www.notesdrive.com In steady state, capacitor acts as open circuit as reactance offered by it to flow of dc (f = 0) is node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docx infinite. As XC = 1 =¥ 2pfC 3 Predict the polarity of the capacitor in the situation described below : A SN B SN Sol. Plate A (+ve) & plate B(–ve) 4. Define ‘quality factor’ of resonance in series LCR circuit. What is its SI unit ? Sol. The Q factor of series resonant circuit is defined as the ratio of the voltage developed across the inductor or capacitor at resonance to the applied voltage, which is the voltage across R. Q = IXL = w0 L IR R It is dimensionless hence, it has no units. OR Quality factor represents sharpness of resonance and is given as Q = wr w2 - w1 it is unitless. 5. An a.c. source of voltage V = V0 sinwt is connected to an ideal inductor. Draw graphs of voltage V and current I versus wt. Sol. V, I I yV V wt O p/2 p 3p 2p wt x I 2 E

ALLENÒ CBSE 123 EXERCISE-II TWO MARK QUESTIONS : 1. A rectangular loop PQMN with movable arm PQ of length 10 cm and resistance 2W is placed in a uniform magnetic field of 0.1 T acting perpendicular to the plane of the loop as is shown in the figure. The resistances of the arms MN, NP and MQ are negligible. Calculate the (i) emf induced in the arm PQ and (ii) current induced in the loop when arm PQ is moved with velocity 20 m/s. ×N× × × × × P× × ×××××××× × × × × × × × ×v ×××××××× × M× × × × × Q × × Sol. (i) Induced emf in arm PQ e = –Blv e = – 0.1 × 10 × 10–2 × 20 e = – 0.2 V (ii) Current induced in loop node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docx I = e = 0.2 = 0.1A www.notesdrive.comR2 EXERCISE-III THREE MARK QUESTIONS : 1. A capacitor (C) and resistor (R) are connected in series with an ac source of voltage of frequency 50 Hz.The potential difference across C and R are respectively 120 V, 90 V, and the current in the circuit is 3 A. Calculate (i) the impedance of the circuit (ii) the value of the inductance, which when connected in series with C and R will make the power factor of the circuit unity. OR The figure shows a series LCR circuit connected to a variable frequency 230 V source. 40W 230V 80 µF 5.0 H (a) Determine the source frequency which drives the circuit in resonance. (b) Calculate the impedance of the circuit and amplitude of current at resonance. (c) Show that potential drop across LC combination is zero at resonating frequency. E

124 Physics ALLENÒ Sol. Given: f = 50 Hz, I = 3A, Vc = 120 V & VR = 90 V (i) Impedance of the circuit: Z= V ÞZ= Vc2 + VR2 I I Z = (120)2 + (90)2 3 Z = 150W (ii) Power factor (cos f) = 1. This is the condition of resonance. Let inductance (L) is connected in series with C and R. At resonance. 40W XL = XC 230V 80 µF XL = 40W Þ L = 40 www.notesdrive.com L= 40 = 40 = 0.4 H node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docx2pf2p × 50p 5.0n H OR (a) At resonance XL = XC Þ wrL = 1 wrC fr = 1 Þ fr = 2p 1 2p LC 5 ´ 80 ´10-6 fr = 1 2p ´ 20 ´10-3 fr = 25 Hz p (b) Impedance of the circuit at resonance Z = R Þ Z = 40W Amplitude of current at resonance Vrms = Irms Z Þ Irms = 230 A 40 Irms = 23 A 4 amplitude, I0 = 2 Irms I0 = 8.1A Potential drop across LC combination VLC = VL – VC = I(XL – XC) At resonance XL = XC Þ XL – XC = 0 VLC = 0 E

ALLENÒ CBSE 125 2. (i) Define self-inductance. Write its SI unit. (ii) A long solenoid with 15 turns per cm has a small loop of area 2.0 cm2 placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 Sol. (i) A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing ? Self inductance of a coil is numerically equal to emf induced in the coil, when rate of change of current is unity. Þ L= - çæè e öø÷ dI dt Or node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docxL = Nf / I www.notesdrive.comIt is equal to the magnetic flux linked with coil when a unit current flow through it. The S.I unit of L is henry (H). (ii) Here, number of turns per unit length, n= N/l = 15 turns/cm=1500 turns/m A = 2 cm2 = 2 × 10–4 m2 dI/dt = (4 – 2)/0.1 or dI/dt = 20 As–1 e = df = d (BA) ëêéQ B = m0 NI ù dt dt l úû e = Ad æ m 0 NI ö = Am0 æ N ö dI dt çè l ø÷ èç l ÷ø dt e = (2 × 10–4) × 4p × 10–7 × 1500 × 20 V e = 7.5 × 10–6 V 3. (i) Define mutual inductance. (ii) A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20A in 0.5 s, what is the change of flux linkage with the other coil? Sol. (i) Mutual Inductance : It is numerically equal to the magnetic flux linked with one coil (secondary coil) when a unit current flows through the other coil (primary coil). (ii) Magnetic flux f1= MI1 As I1 = 0 \\ f1= 0 f2 = MI2 = 1.5 × 20 = 30 Change in flux df = f2 - f1 = 30 – 0 = 30 weber. E

126 Physics ALLENÒ 4. Draw a schematic arrangement for winding of primary and secondary coils in a transformer with the two coils on separate limbs of the core. State its underlying principle and find the relation between the primary and secondary voltages in terms of the number of turns of the primary and secondary windings. How are the currents in the primary and secondary coils related to the voltages in the case of an ideal transformer ? Sol. Laminated Copper core winding Secondary Primary NP NS Copperwww.notesdrive.com NS< Np winding node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docx Principle of a transformer : When alternating current flows through the primary coil, an emf is induced in the neighbouring (secondary) coil due to mutual induction. Induced emf produced in the primary coil is EP = – NP (df/dt) (self induction) In the secondary coil, it will be ES = –NS (df/dt) (mutual induction) Hence ES = -NS df = NS EP -NP dt NP df dt For ideal transformer ES = VS = NS EP VP NP Let IP and IS be the current in the primary and secondary coil respectively, under ideal condition. Instantenous power input is equal to instantaneous power output VPIP = VSIS or VS = IP = NS VP IS NP 5. Define mutual inductance between a pair of coils. Derive an expression for the mutual inductance of two long coaxial solenoids of same length wound one over the other. (Refer to Ch.-6, NCERT Q. No. 8) OR Define self-inductance of a coil. Obtain the expression for the energy stored in an inductor L connected across a source of emf. (Refer to Ch.-6, NCERT Q. No. 9) E

ALLENÒ CBSE 127 6. (i) When an AC source is connected to an ideal capacitor, show that the average power supplied by the source over a complete cycle is zero (ii) A bulb is connected in series with a variable capacitor and an A.C. source as shown. What happens to the brightness of the bulb when the key is plugged in and capacitance of the capacitor is gradually reduced ? Bulb C () ~ Sol. (i) Given V = V0 sin wt q = CV q = C V0 sin wt +– node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docx ~ www.notesdrive.com V = V0 sin wt dq = CV0 d sin wt dt dt i = CV0 w(cos wt) i = æ V0 cos wt Þ i = V0 sin (wt + p / 2) 1 ö Xc çè wC ÷ø Here XC = 1 and I0 = VO wC XC Average power T òPav = Vi dt 0 ò= V02 T + p / 2)dt XC (sin wt)(sin wt 0 ò= V02 T XC (sin wt)(cos wt)dt 0 V02 T ìïT sin(2wt)dt 0 2XC í sin(2wt)dt 0 ò ò= = ïî 0 =0 (ii) When AC source is connected, the capacitor offers capacitive reactance XC = 1/Cw. The current flows in the circuit and the lamp glows. On reducing C, XC increases and current reduces, Therefore, glow of the bulb reduces. E

128 Physics ALLENÒ 7. A capacitor of unknown capacitance, a resistor of 100 W and an inductor of self inductance L = (4/p2) henry are connected in series to an ac source of 200 V and 50 Hz. Calculate the value of the capacitance and impedance of the circuit when the current is in phase with the voltage. Calculate the power dissipated in the circuit. Sol. Current in phase with voltage means angle between emf and current is zero. This is resonance condition, So Inductive reactance = Capacitive reactance Þ XL = XC Þ wL = 1 ÞC= 1 , given L = 4/p2 f = 50Hz E = 200V wC w2L So C = 1´ p2 (50)2 Þ C = 25 µF (2p)2 ´4´ Power dissipatedwww.notesdrive.com P = V2/R = 200 ´ 200 100 node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docx P = 400W 8. (i) When an AC source is connected to an ideal inductor show that the average power supplied by the source over a complete cycle is zero. lamp (ii) A lamp is connected in series with an inductor and an AC source. What happens to the brightness of the lamp when the key is plugged in and an iron rod is inserted inside the inductor ? Explain. () ~ Sol. (i) Given V = V0 sin wt V =L di Þ di = V dt L dt L di = sin wt dt By integration i = -V0 cos wt ~ wL V = V0 sin wt V0 \\ i= - wL sin é p - wt ù Þ I0 sin é wt - p ù êë 2 úû êë 2 úû where I0 = V0 wL Average power T òPav = Vi dt o ò ò ò= V02 T sin wt cos wt dt V02 T wt) dt ìïT sin(2wt)dt 0 0 - wL o Þ - 2wL o sin(2 Þ í = Þ ïî 0 (ii) When an iron rod is inserted into the inductor, the self-inductance of the inductor increases. On increasing L, XL increases and current reduces. Therefore glow of the bulb reduces. E

ALLENÒ CBSE 129 EXERCISE-IV FIVE MARK QUESTIONS : 1. Derive the expression for the magnetic energy stored in an inductor when a current I develops in it. Hence, obtain the expression for the magnetic energy density. (Refer to NCERT Q. No. 9) Energy stored in an inductor: Work has to be done by battery against the opposing induced emf in establishing a current in an inductor. The energy supplied by the battery is stored in the inductor. 2. A device 'X' is connected to an ac source V = V0 sin wt. The variation of voltage, current and power in one cycle is shown in the following graph : YA B C node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docx 0 p 2p wt www.notesdrive.com (a) Identify the device 'X'. (b) Which of the curves A, B and C represent the voltage, current and the power consumed in the circuit? Justify your answer. (c) How does its impedance vary with frequency of the ac source? Show graphically. (d) Obtain an expression for the current in the circuit and its phase relation with ac voltage. OR (a) Draw a labelled diagram of an ac generator. Obtain the expression for the emf induced in the rotating coil of N turns each of cross-sectional area A, in the presence of a magnetic field Br . (b) A horizontal conducting rod 10 m long extending from east to west is falling with a speed 5.0 ms–1 at right angles to the horizontal component of the Earth's magnetic field, 0.3 × 10–4 Wb m–2. Find the instantaneous value of the emf induced in the rod. Sol. (a) Capacitor (b) Curve – A represent power, because in pure capacitive ac circuit, power consumed in one cycle is zero and frequency of power is twice the frequency of voltage (or current). Curve-B represent voltage. Curve-C represent current because in pure capacitive ac circuit, current leads the voltage by p/2. (c) Z= XC Þ Z= 1 2p f Z f E

130 Physics ALLENÒ (d) We know, V = q or q = CV C C Also, q = CV0sinwt V = V0 sin wt or dq = CV0 d (sin wt ) dt dt I = CV0 (w coswt) I= V0 cos wt æ 1 ö çè wC ÷ø I = V0 cos wt Q XC = 1 XC wC I = I0sin (wt + p/2) Q I0 = V0 www.notesdrive.com XC \\ Clearly, current leads by (p/2) with voltage. node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docx OR (a) w Axis Coil B N qS A B1 S1 S2 B2 R Load B1 & B2 are carbon brushes. S1 & S2 are slip rings. Let at any instant (t), the angle between Br and Ar is q, then the magnetic flux linked with coil at that instant. N = Number of turns in the coil ( )fB = N Br · Ar = NBA cos q A = Area of the coil w = Angular velocity of coil fB = NBA cos wt {Q q = wt} dfB = NBA w (– sin wt) dt Thus induced emf, e = – dfB dt or e = NBAw sinwt e = e0 sin wt {Q e0 = NBAw} E

ALLENÒ CBSE 131 (b) Given : L = 10 m, v = 5 m/s, BH = 0.3 × 10–4 Weber Induced emf (e) = BHvL = 0.3 × 10–4 × 5 × 10 = 15 × 10–4 volt = 1.5 mV 3. (a) Explain the meaning of the term mutual inductance. Consider two concentric circular coils, one of radius r1 and the other of radius r2 (r1 < r2) placed coaxially with centres coinciding with each other. Obtain the expression for the mutual inductance of the arrangement. (Refer to Ch.-6, NCERT Q. No. 12) (b) A rectangular coil of area A, having number of turns N is rotated at ‘f’ revolutions per second in a uniform magnetic field B, the field being perpendicular to the coil. Prove that node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docx the maximum emf induced in the coil is 2 pf NBA. www.notesdrive.com Sol. (b) Magnetic flux linked with coil = f = NBAcosq The induced emf, e = –df/dt çæèQ w = dq ö dt ÷ø e = - df = æ -NBA(- sin q) dq ö = NBA sin q(2pf) dt èç dt ø÷ For maximum induced emf, sinq = 1 \\ e = NBA(2pf) Þ e = 2pfNBA 4. (i) Draw a labelled diagram of a step-down transformer. State the principle of its working (ii) Express the turn ratio in terms of voltages. (iii) Find the ratio of primary and secondary currents in terms of turn ratio in an ideal transformer. (iv) How much current is drawn by the primary of a transformer connected to 220 V supply when it delivers power to a 110 V – 550 W refrigerator ? Sol. (i) Np Ns Secondary Primary Ns < Np Principle: Transformer works on principle of mutual induction , in which an EMF is induced in the secondary coil by the change in magnetic flux in primary coil. Working : When alternating current source is connected to the ends of primary coil, the current changes continuously in the primary coil, due to which magnetic flux linked with the secondary coil changes continuously. Therefore, the alternating emf of same frequency is developed across the secondary terminals. E

132 Physics ALLENÒ (ii) NS = VS ì NS = turn ratio NP VP í NP î (iii) For ideal transformer Output power = Input power VSIS = VPIP VS = IP ì VS = NS VP IS í VP NP î NS = IP NP IS (iv) Given VP = 220 V VS = 110 V P = 550 W www.notesdrive.com IP = ? node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docx IP = Power = P Primary Voltage VP IP = 550 = 2.5 A 220 5. A 2 µF capacitor, 100 W resistor and 8 H inductor are connected in series with an AC source. (i) What should be the frequency of the source such that current drawn in the circuit is maximum, What is this frequency called ? (ii) If the peak value of e.m.f. of the source is 200 V, find the maximum current. (iii) Draw a graph showing variation of amplitude of circuit current with changing frequency of applied voltage in a series LCR circuit for two different values of resistance R1 and R2 (R1 > R2). (iv) Define the term ‘Sharpness of Resonance’. Under what condition, does a circuit become more selective ? Sol. (i) Source frequency, when current is maximum is given by 11 {L = 8H and C = 2µF f= = 2p LC 2p 8´2´10-6 f= 1 2p´ 4´10-3 f = 39.80 Hz The frequency at which current maximum, is called resonant frequency. (ii) given E0 = 200V, R = 100W Imax = E0 200 = 2A R = 100 E

ALLENÒ CBSE 133 (iii) R2 I R1 w0 w (iv) The sharpness of resonance is given by quality factor (Q factor) of a resonant ckt. It is defined as the ratio of the voltage drop across inductor (or capacitor) to the applied voltage. Sharper the curve, ckt will be more selective. For resistant R2 , circuit is more selective. 6. (i) An a.c. source of voltage V = V0sinwt is connected to a series combination of L, C and R. Use the phasor diagram to obtain expressions for impedance of the circuit and phase angle between voltage and current. Find the condition when current will be in phase with the node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docx voltage. In this condition what we call this circuit ? www.notesdrive.com (ii) In a series LR circuit XL = R and power factor of the circuit is P1. When capacitor with capacitance C such that XL = XC is put in series, the power factor becomes P2. Calculate P1/P2. Sol. (i) Applied ac voltage V = V0 sin wt VC VR VL L CR V = V0 sin wt Let I be the rms value of current flowing through all thYe circuit elements, then VL B VL = IXL , VC = IXC and VR = IR D Let VC > VC . E x The resultant of VR and (VL – VC) VL – VC is given by O V VL – VC VC f V = ( )VR2 + VL - VC 2 VR A I V = I2R2 + (IXL - IXC )2 C V = R 2 + ( X L - X C )2 –Y I But V = Z, where Z is the impedance of seires LCR circuit. I \\ Impedance of LCR circuit is given by Z = R2 + ( XL - XC )2 E

134 Physics ALLENÒ Phase angle between voltage and current is f, then tan f = VL - VC Þ tan f = IX L - IX C VR IR f = tan -1 æ XL - XC ö çè R ÷ø Equation of current is given by i = i0 sin (wt - f) If XL = XC then f = 0. In this case current will be in phase with the voltage. In this condition the circuit is called resonant circuit. (ii) In series LCR circuit impedance Z = R2 + ( X L - X C )2 and power factor, P = R/Z www.notesdrive.com Case I :- In LR ckt When XL = R, So Z = (2R2)½ node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docx Z= 2R Þ Now P1 = R = R Þ P1 = 1 Z 2R 2 Case II :- XL = XC, Z = R, So Power factor P2 becomes equal to 1 P2 = 1 So ratio P1 = 1 : 1 Þ P1 = 1 P2 2 1 P2 2 7. In a series LCR circuit connected to an a.c. source of voltage V = Vm sinwt, use phasor diagram to derive an expression for the current in the circuit. Hence obtain the expression for the power dissipated in the circuit. Show that power dissipated at resonance is maximum. Sol. Suppose OA, OB, OC represents the magnitude of phasor VR, VL and VC respectively. In case of VL > VC, the resultant of (VR) and (VL–VC), represented by OE. Thus from DOAE OE = OA2 + AE2 V = ( )VR2 + VL - VC 2 VL Substituting the value of VR, VL and VC we have B EV V = (IR )2 + (IX L - IX C )2 VL – VC or I = V Of A VR I (R)2 + (XL - XC )2 VC C The effective opposition offered by L, C, R to a.c. supply is called impedance of LCR circuit and represented by Z. I V =Z Z = R2 + æ wL - 1 ö2 èç wC ø÷ E

ALLENÒ CBSE 135 Also from DOAE tanf = AE = VL - VC OA VR or tanf = ( X L - X C ) R or f= tan-1 ( XL - XC ) R Power dissipation in LCR circuit : The instantaneous power supplied by the source is P = VI P = (Vmsinwt) × imsin(wt + f) = Vm im [cos f - cos(2wt + f)] 2 [2sinAsinB = cos(A–B) – cos(A+B)] node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docx For average power the second term becomes zero in the complete cycle. www.notesdrive.com So Pav = Vm i m cos f 2 Pav = Vm im cos f = Vrms irms cosf 2 2 Pav = Vrmsi rms cos f At resonance condition cosf = 1 (because f = 0), so R becomes effective impedance of a circuit. So power dissipated is maximum at resonance condition. 8. (a) When a bar magnet is pushed towards (or away) from the coil connected to a galvanometer, the pointer in the galvanometer deflects. Identify the phenomenon causing this deflection and write the factors on which the amount and direction of the deflection depends. State the laws describing this phenomenon. (b) Sketch the change in flux, emf and force when a conducting rod PQ of resistance R and length l moves freely to and fro between A and C with speed v on a rectangular conductor placed in uniform magnetic field as shown in the figure. • • •P • C B• • • • v •••• • •A •Q • x=0 x=b x=2b Sol. (a) When a bar magnet pushed towards or away from coil, magnetic flux passing through coil change with time and cause induced emf. This phenomenon is called EMI. Induced emf in the coil is given as e = -N df dt Direction & amount of deflection depend on the motion of magnet whether it is moving towards coil or away from the coil. E

136 Physics ALLENÒ Farady’s law of electromagnetic induction :- Whenever there is a change in magnetic flux linked with a coil, an emf is induced in the coil. The induced emf is proportional to the rate of change of magnetic flux linked with the coil. e µ df dt e µ -df Þ e = -K df dt dt Where K is constant and negative sign represents opposition to change in flux. In SI system f is in weber, t in seconds, e in volt & K = 1 \\ e = - df dt www.notesdrive.com If the coil has N turns, then node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docx e = -N df dt (b) Case I : When PQ moves forward (i) For 0 < x < b Magnetic field, B exists in the region \\ Area of loop PQRS = lx \\ Magnetic flux linked with loop PQRS, f = BA = Blx f = Blx ....(i) [b > x > 0] (ii) For 2b > x > b B=0 \\ Flux linked with loop PQRS is uniform and given by f¢ = Blb (x = b) Forward journey Thus, for 2b > x > b Flux, f’ = Bbl [constant] Return journey For b < x < 2b Flux, f' = constant = Bbl For 0 < x < b, f = Blx [Decreasing] E

ALLENÒ CBSE 137 Graphical representaiton Forward Return journey f journey f = Bbl 0 b 2b b 0 x Case II For b>x>0 As f = Blx Þ df = Bldx = Bvl êëéQ v = dx ù dt dt dt ûú node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docx Induced emf, e = –df/dt = –vBl Þ df' =0 www.notesdrive.com dt For 2b > x > b As f¢ = Bbl Þ e = 0 Forward journey, Forward Retrurn For b > x > 0 journey journey e = –vBl, +Blv For 2b > x > b, e = 0 emf b Return journey, –Blv 2b b 0 x For b > x > 0 e = vBl For 2b > x > b, Variation of induced emf e=0 OR ® Refer NCERT solved example 6.8 (Page No. 216) 9. (a) State the principle of an ac generator and explain its working with the help of a labelled diagram. Obtain the expression for the emf induced in a coil having N turns each of cross- sectional area A, rotating with a constant angular speed 'w' in a magnetic field Br , directed perpendicular to the axis of rotation. (Refer to 5 Marks, PYQ.-2) (b) An aeroplane is flying horizontally from west with a velocity of 900 km/hour. Calculate the potential difference developed between the ends of it's wings having a span of 20m. The horizontal component of the Earth's magnetic field is 5×10–4 T and the angle of dip is 30°. E

138 Physics ALLENÒ OR A device X is connected across an ac source of voltage V = V0 sin wt. The current through X is given as I = I0 sin æ wt + p ö . çè 2 ÷ø (a) Identify the device X and write the expression for it's reactance. (b) Draw graphs showing variation of voltage and current with time over one cycle of ac, for X. (c) How does the reactance of the device X vary with frequency of the ac ? Show this variation graphically. (d) Draw the phasor diagram for the device X. Sol. (b) Potential difference developed between the ends of the wings 'e' = Blv (a) Given Velocity (v) = 900 km/hour = 250 m/s www.notesdrive.com Wing span (l) = 20 m node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docx Vertical component of Earth's magnetic field BV = BH tan d = 5 × 10–4 (tan 30º) tesla \\ Potential difference = 5 × 10–4 (tan 30º) × 20 × 250 = 5 ´ 20 ´ 250 ´10-4 3 = 1.44 volt OR X : capacitor Reactance XC = 1 = 1 C wC 2pf v i (b) 0 t1 2t (c) Reactance of the capacitor varies in inverse proportion to the frequency i.e., XC µ 1 f XC V0 sin wt1 V f t1 I imsin ( t1 + /2) E

ALLENÒ CBSE 139 10. The emf induced across the ends of a conductor due to its motion in a magnetic field is called motional emf. It is produced due to the magnetic Lorentz force acting on the free electrons of the conductor. For a circuit shown in figure, if a conductor of length l moves with velocity v in a magnetic field B perpendicular to both its length and the direction of the magnetic field, then all the induced parameters are possible in the circuit. (i) Direction of current induced in a wire moving in a magnetic field is found using :- (A) Fleming's left hand rule (B) Fleming's right hand rule node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docx(C) Ampere's rule(D) Right hand clasp rule www.notesdrive.comAns. B (ii) Bicycle generator creates 1.5 V at 15 km/hr.The EMF generated at 10 km/hr is :- (A) 1.5 volts (B) 2volts (C) 0.5volts (D) 1 volt Ans. D (iii) A 0.1 m long conductor carrying a current of 50 A is held perpendicular to a magnetic field of 1.25 mT. The mechanical power required to move the conductor with a speed of 1 m s–1 is :- (A) 62.5 mW (B) 625 mW (C) 6.25 mW (D) 12.5 mW Ans. C (iv) A conducting rod of length / is moving in a transverse magnetic field of strength B with velocity V. The resistance of the rod is R. The current in the rod is :- (A) BVl (B) Zero (C) BVl (D) B2V2l2 R R Ans. C 11. The power averaged over one full cycle of a.c. is known as average power. It is also known as true power. Pav = VrmsIrms cos j = Vo Io cos j 2 Root mean square or simply rms watts refer to continuous power. A circuit containing a 80 mH inductor and a 60 µF capacitor in series is connected to a 230V, 50 Hz supply. The resistance of the circuit is negligible. (i) The average power transferred to inductor is (A) Zero (B) 7W (C) 3W (D) 5W Ans. (A) E

140 Physics ALLENÒ (ii) What is the total average power absorbed by the circuit? (A) 10VV (B) Zero (C) 2.5W (D) 15W (D) 6.33A Ans. (B) (D) 7.52A (D) Zero (iii) The value of current amplitude is (A) 15A (B) 11.63A (C) 17.63A Ans. (B) (iv) Find rms value. (A) 6A (B) 5.25A (C) 8.23A Ans. (C) (v) The average power transferred to the capacitor is (A) 5 (B) 11 W (C) 15W Ans. (D) www.notesdrive.com node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docxEXERCISE - V (RACE) CHAPTER-6 : ELECTROMAGNETIC INDUCTION For question numbers 1 to 5 two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below. (a) Both A and R are true and R is the correct explanation of A (b) Both A and R are true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false and R is also false 1. Assertion : At any instant, if the current in an inductor is zero, then the induced emf may not be zero. Reason : An inductor tends to keep the flux (i.e. current) constant. (i) a (ii) b (iii) c (iv) d 2. Assertion : The quantity L/R possesses dimensions of time. Reason : To reduce the rate of growth of current through a solenoid, we should increases the time constant (L/R). (i) a (ii) b (iii) c (iv) d 3. Assertion : Self inductance is called the inertia of electricity. Reason : The inductor opposes change in current. (i) a (ii) b (iii) c (iv) d 4. Assertion : A small magnet takes longer time in falling in a hollow metallic tube without touching the wall. Reason : There is a opposition of motion due to production of eddy currents in metallic tube. (i) a (ii) b (iii) c (iv) d 5. Assertion : Farady's laws are conequences of conservation of energy. Reason : In a purely resistive AC circuit, the current lags behind the emf in phase. (i) a (ii) b (iii) c (iv) d E

ALLENÒ CBSE 141 6. Give one use of eddy currents. 7. Write SI unit of self inductance. 8. Write the Lenz' law. 9. Why, the motion of copper plate is damped when it is allowed to oscillate between the two poles of a magnet. 10. State Farady's law of EMI. 11. A circular loop is moved through the region of uniform magnetic field. Find the direction of induced current (clock wise or antic lock wise) when the loop moves (i) into the e field, and (ii) out of the field. 12. A closed loop moves normal to the constant electric field between the plates of a large capacitor. Is a current induced in the loop (i) when it is wholly inside the region between the capacitor plates, and (ii) when it is partially outside the plates of the capacitor? The electric field is normal to the plane of the loop. 13. When a coil is rotated in a uniform magnetic field at constant angular velocity, will the magnitude of induced emf set up in the coil be constant? Why? 14. The closed loop (PQRS) of wire is moved into a uniform magnetic field at right angles to the plane of the paper as shown in the figure. Predict the direction of the induced current in the loop. 15. A light metal disc on the top of an electromagnet is thrown up as the current is switched on. Why? Give reason. RACE CHAPTER-6 (SOLUTIONS) node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docx 1. (i) 2. (ii) 3. (i) 4. (i) 5. (iii) www.notesdrive.com 6. Eddy currents are used in inductance furnace. Strong eddy currents produce desired heating. 7. Henry (H) 8. The direction of induced emf is such that it tends to produce a current, which opposes the change in magnetic flux that induced it. 9. Damping due to eddy currents. 10. The magnitude of emf induced in a circuit is directly proportional to the rate of change of magnetic flux linked with the circuit. e = - df dt 11. (i) Anticlockwise, (ii) Clockwise 12. No current is induced in either case. Current cannot be induced by changing the electric flux. 13. No, the induced emf will vary with time and will be sinusoidal due to the change in orientation of the coil w.r.t. the magnetic field. 14. Anticlockwise. 15. Due to the eddy currents. As these currents produce opposite polarity on the lower side of the disc. E

142 Physics ALLENÒ EXERCISE - V (RACE) CHAPTER-7 : ALTERNATING CURRENT For question numbers 1 to 5 two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below. (a) Both A and R are true and R is the correct explanation of A (b) Both A and R are true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false and R is also false 1. Assertion : Average power loss in series LC circuit is always zero. Reason : Average value of voltage and current in A.C. is zero. (i) a (ii) b (iii) c (iv) d www.notesdrive.com 2. Assertion : Capacitor serves as a block for D.C, offers an easy path to AC. node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docx Reason : Capacitive reactance is inversely proportional to frequency. (i) a (ii) b (iii) c (iv) d 3. Assertion : When capacitve reactance is smaller then the inductive reactance in series LCR circuit, voltage leads the current. Reason : In series LCR circuit inductive reactance is always greater then capacitive reactance. (i) a (ii) b (iii) c (iv) d 4. Assertion : In the series LCR circuit, the impedance is minimum at resonance. Reason : The currents in inductor & capacitor are same in series LCR circuit. (i) a (ii) b (iii) c (iv) d 5. Assertion : In series LCR circuit phase difference between current & voltage is never zero. Reason : Voltage and current are never in phase. (i) a (ii) b (iii) c (iv) d 6. Define root mean square of AC current. 7. An ordianary moving coil ammeter used for d.c. cannot be used to measure a.c. even if its frequency is low, why? 8. A capacitor is 50 pf is connected to an A.C. source of frequency 1 KHz. Calculate its reactance. 9. Write the principle of transformer. 10. Define the term sharpness of resonance. 11. How can you improve the quality factor a series resonance circuit ? 12. The electric mains [220V, 50 Hz ] are connected to a resistor of 50 KW . Find the current as a function of time & the effective current. 13. Can ever the rms value be equal to the peak value of an ac? 14. When an alternating current is passed through a moving coil galvanometer, it shows no deflection. Why? 15. A 100 V source of frequency 500 Hz is connected to an LCR series circuit with L= 8.1 mH , C = 12.5 µf, and R = 10W. Find the p.d. across the resistance. E

ALLENÒ CBSE 143 RACE CHAPTER-7 (SOLUTIONS) 1. (ii) 2. (i) 3. (iii) 4. (ii) 5. (iv) 6. It is the value of DC current which would produce same heat in given resistance in given time as it done by the alternating when passed through the same resistance for the same time. node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docx7. The average value of AC current for a cycle is zero. So a DC ammeter will always read zero in www.notesdrive.comAC. circuit. 8. XC = 1 = 2p´103 1 = 107 W . Cw ´ 50 ´10-12 p 9. Transformer works on the principle of mutual inductance. 10. Sharpness of resonance (Quality factor) of a resonant circuit is defined as the ratio of the voltage drops across inductor (or capacitor) to the applied voltage. Q = wL = 1 R CwR 11. To improve quality factor, ohmic resistance should be made as small as possible. 12. Given :Vrms = 220 V, \\ Irms = Vrms = 220 = 4.4mA R 50 ´103 I0 = 2 Irms = 6.22mA and I = 6.22 ´10-3 sin100pt 13. Yes, when the ac is a square wave. 14. A moving coil galvanometer measures an average value of current, which is zero for ac. Hence, no deflection is shown by galvanometer. 15. As per given data circuit is found under resonant condition (XC = XL) é XC = 1 = 25.47Wùú Þ VR = 100 V ê = ú ê 12.5´10-6 ´ 2p´ 500 ëêXL 8.1´10-3 ´ 2p´ 500 = 25.47W úû E

144 Physics ALLENÒ EXERCISE - VI (MOCK TEST) CHAPTER-6 : ELECTROMAGNETIC INDUCTION 1. A wire of length 0.3 m moves with a speed of 20m/s perpendicular to the magnetic field of strength 1T. Calculate the induced emf . [1] 2 Predict the polarity of the capacitor in the situation described below : [1] SN A B SN 3. When a coil is rotated in a uniform magnetic field at constant angular velocity, will the magnitude of induced emf set up in the coil be constant? Why?www.notesdrive.com [1] 4. Define self-inductance. Write its SI units. node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-1\\ 05_Unit-4.docx[2] 5. Obtain the expression for the energy stored in an inductor L connected across a source of emf.[2] 6. How is the mutual inductance of a pair of coils affected when (1) Separation between the coils is increased. (2) The number of turns of each coil is increased. [2] 7. Why is the emf zero, when maximum number of magnetic lines of force pass through the coil?[2] 8. A rectangular loop PQMN with movable arm PQ of length 10 cm and resistance 2W is placed in a uniform magnetic field of 0.1 T acting perpendicular to the plane of the loop as is shown in the figure. The resistances of the arms MN, NP and MQ are negligible. Calculate the (i) emf induced in the arm PQ and (ii) current induced in the loop when arm PQ is moved with velocity 20 m/s. [3] 9. A circular copper disc. 10 cm in radius rotates at a speed of 2p rad/s about an axis through its centre and perpendicular to the disc. A uniform magnetic field of 0.2T acts perpendicular to the disc. (1) Calculate the potential difference developed between the axis of the disc and the rim. (2) What is the induced current if the resistant of the disc is 2W ? [3] 10. A long solenoid with 15 turns per cm has a small loop of area 2.0 cm2 placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing? [3] E