Electromagnetic Induction-09-OBJECTIVE & SUBJECTIVE LEVEL – I

OBJ. LEVEL – I 1. From dimensional homogeneity Weber Ans. (b) 2. Ans. (c) 3. As from figure it is clear that no charge particle will ever complete the rotation. Ans. (a) 4. As does not mean that or vice-versa If , may or may not be zero Ans. (c) 5. As time varying magnetic field can produce induce electric field which set up current in the ring. But a current carrying coil placed in uniform magnetic field will not experience any force. Also magnetic moment and area vectors are parallel to each other, so that torque about any axis is zero. Because of current element in magnetic field, the ring has a tension force along its length. Ans. (d) 6. As , if is parallel to , there will be no force. Also in (b) and (c) case is to , so the direction of force is to and it is not possible to drift electron in that direction results no induced emf Ans. (d) 7. As current in xy wire increases, so flux across both the loops increases, so induced must oppose external magnetic field. As also due to wire in loop is into and in loop A is outward so induced must be outward in loop B and inward in loop A. So the direction of induced current in B is anticlockwise and in A, it is clockwise. Ans. (a) 8. From Lenz’s law, emf will induced in the ring in such a way that it always oppose the change in flux i.e., in both the motion the ring’s polarity will develop in such a way that it always oppose the change in flux across ring and so magnet will get repelled while approaching and attracted while receding. So, the result of both the motion, the magnet’s acceleration will be less than g. Ans. (c) 9. In L-R circuit, current decreases with time as at Ans. (b) 10. 11. Ans. (d) 12. Ans. (c) 13. As Ans. (d) 15. As disc is conducting, so any free electron which has velocity r tangential will experience a force away from centre and the centre is positively charged and periphery gets negatively charge. So, potential difference between C and rim is i.e., Ans. (c) SOLUTION OBJ. LEVEL - II 1. From, magnetic force will act on electron towards A and because of this A gets positively charged and D has negative charge Ans. (d) 2. In position II, there is no change in flux and so emf induced in this case is zero. Ans. (a) 3. Motional emf for each wire = Volt In parallel, [As r1 = r2 = 2 ] Ans. (b) 4. As flux across ring is constant so there is no induced emf across the ring and hence no current. Also as CE is parallel to , there is no potential difference across C and E. Ans. (c) 5. Ans. (c) 6. emf Ans. (d) 7. For bigger ring highest point is at high potential w.r.t. bottom point and for smaller ring highest point is at low potential w.r.t. bottom point. Thus As Ans. (d) 8. Ans. (a) 9. Ans. (b) 10. Time spent 11. The duration of induced emf is the duration in which flux is changed i.e., 10-3 s. 12. At steady state current in the coil As flux is constant, Ans. (c) 13. At steady state i.e. for a long time after switch S is closed current in the branch of inductor is and after opening the switch at t = 0, energy stored in the inductor must be same and have same current will flow across inductor thus p.d. across it in at Ans. (d) 14. As Ans. (c) 15. As Keeping one resistance at rest, Ans. (d). SOLUTION SUB. LEVEL - I(C.B.S.E.) 1. According to Lenz’s law, as flux changes across any loop, induced emf will set-up in such a way that it opposes the change in flux. i.e., If is +ve, induced is opposite to and if is –ve, induced will support In figure (i) is +ve, so will oppose i.e., for dot current must be anticlockwise But for figure (ii) and (iii) as is –ve, so will support i.e. for to be current must be clockwise. 2. From Lenz’s law (a) As magnet approaches coil, south pole will develop in the near and of the coil and thus a current p to q set-up across the coil (b) along q P, along xy (c) along x y z (d) along z y x (e) along x y (f) no induced current 3. (a) As area of the loop increases, flux also increases so induced will oppose . Hence to make onward current induced in the coil is anticlockwise. (b) Above explanation, current is anticlockwise. 4. 5. As flux across the loop, Volt 6. V. 7. Let  be the angle area vector of the coil makes with magnetic field, the flux linked across the coil at that instant, emf induced V Heat dissipated Source of this power is the mechanical source which is rotating the coil. 8. emf induced 9. 10. Magnetic field inside circular loop 11. 12. 13. 14. (a) (b) 15. Flux across disc As B is switched off Again . SOLUTION SUB. LEVEL - II 1. emf induced Total resistance induced current 2. As 3. Flux associated with the loop 4. To lift the ‘wire frame’ . 5.(a) As B increases with time, so induced electric field is set-up in such a way that it will produce magnetic field outward. Here lines are concentric circle with anticlockwise sense. (b) Again (c) From symmetry, we can say that induced emf across BC is same as that of AB. (d) For side CA, from symmetry about point O, (e) From faraday’s law, emf induced across loop, (f) Adding results of part (b), (c) and (d), we get The result is same as that of part (e) 6.(a) Let I be the current in bigger coil, Flow across smaller coil Mutual inductance, (b) emf induced in, B (c) 7.(a) Let x be the position of conductor from resistance R and v be the instantaneous velocity. emf induced = B l. v current, Again, 8. Time taken by the loop  complete a solution . After time the loop will be 11th to magnetic field. Let at any time t the loop is inside the field and making angle  with the field Angle between the area normal areal field Flux through the loop 9. T/s emf induced across loop, current induced in the loop A 10. L = 20 mH, R = 10 , V = 5.0 V As emf induced, Rate of charge of induced emf, (a) (b) (c) . SOLUTION SUB. LEVEL - III 1. cos  t cos  t Amplitude of the current = 0.5 A 2. As magnetic filed varies with time, induced electric field is given by or Taking a differential element of are h.dr and applying Ohm’s law, we get or 3. At any time ‘t’ Area of the sector, Flux linked across semi-circle Taking anticlockwise as –ve and clockwise +ve, we have 4. After a long time, the current in the circuit will be and so the flux associated with coil As inductance was decreased abruptly, the flux linked across the coil is constant i.e., i.e., at t = 0 From Kirchhoff’s voltage law, or or 5. The circuit can be redrawn as Here emf induced across OA, Current in rode OA, Also 6.(a) Let v be the speed of the frame at any instant. emf induced = current in the wire, Force due to current on the frame Applying Newton’s 2nd law (b) Again or or for (c) Again, 7.(a) Induced emf across the connector induced current, (b) Force required to maintain the constant speed V, 8. As loop starts entering the region containing B field, flux across loop, Induced current, Time in one revolution in which induced current will be present in the loop heat generated in a revolution rad/s 9. Again energy stored inside the rod 10. From equilibrium condition SOLUTION SUB. LEVEL 1.(a) Here torque must balance gravity. Now, Thus m is negative I should be clockwise (b) (c) 2. For the circuit shown taking Kirchoffs law or considering voltage (a) (b) Net charge which flows through resistance (c) We know that …(i) for When Sub. in 1 3.(a) be the current flown through the wire is the magnetic field. …(i) where Area (b) Since and = angular acceleration [From perpendicular axis theorem ] Now For given value of all the quantities we have 4. Let m = mass of bar L = 0.2 kg When x = length of the rail V = terminal velocity of the bar e = emf induced in the upper circuit = emf induced in the lower circuit = B/V i = current flowing through the rod Fmag = magnetic force acting on the rod = I l B According to the Lenz’s law, the direction of current in the two circuits are shown in figure. When terminal velocity is attained. i/B = mg. . Now applying Kirchhoff’s law to the two circuit, we get For the circuit ACEFC, e = i1R1 …(i) For the circuit BDEFD, e = i2R2 …(ii) At the point E, …(iii) Multiplying both sides of equation (iii) by e, we get By (i) and (ii) Given that power dissipated in R1 and R2 are 0.76 W and 1.2 W respectively Terminal velocity, Again power dissipated in R1 = power dissipated in R2 = By equation (i), By equation (ii), 5. Flux through the square loop Induced emf Charge on the capacitor (say) Current in the loop