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Chapter 6 Electromagnetic Induction Solutions

Question - 11 : - Suppose the loop in Exercise 6.4 is stationary but the current feedingthe electromagnet
that produces the magnetic field is gradually reduced so that the fielddecreases from its initial value of 0.3 T at the rate of 0.02 T s-1.If the cut is joined and the loop has a resistance of 1.6 Ω, how much power isdissipated by the loop as heat ? What is the source of this power ?

Answer - 11 : -


Source of this power is the external agencywhich brings change in magnetic field.

Question - 12 : - A square loop of side 12 cm with its sides parallel to X and Y axes ismoved with a velocity of 8 cm s_1 in the positive x-directionin an environment containing a magnetic field in the positive z-direction. Thefield is neither uniform in space nor constant in time. It has a gradient of 10-3 Tcm-1 along the negative x- direction (that is it increases by
10-3 T cm-1 as one moves in the negativex-direction), and it is decreasing in time at the rate of 10-3 Ts_1. Determine the direction and magnitude of the induced current inthe loop if its resistance is 4.50 mΩ.

Answer - 12 : -


Direction of induced current is such that itincrease the magnetic flux linking with the loop in positive 2-direction.

Question - 13 : - A square loop of side 12 cm with its sides parallel to X and Y axes ismoved with a velocity of 8 cm s_1 in the positive x-directionin an environment containing a magnetic field in the positive z-direction. Thefield is neither uniform in space nor constant in time. It has a gradient of 10-3 Tcm-1 along the negative x- direction (that is it increases by
10-3 T cm-1 as one moves in the negativex-direction), and it is decreasing in time at the rate of 10-3 Ts_1. Determine the direction and magnitude of the induced current inthe loop if its resistance is 4.50 mΩ.

Answer - 13 : -


Direction of induced current is such that itincrease the magnetic flux linking with the loop in positive 2-direction.

Question - 14 : - It is desired to measure the magnitude of field between the poles of apowerful loud speaker magnet. A small flat search coil of area 2 cm2 with25 closely wound turns, is positioned normal to the field direction, and thenquickly snatched out of the field region. Equivalently, one can give it a quick90° turn to bring its plane parallel to the field direction. The total chargeflown in the coil (measured by a ballistic galvanometer connected to coil) is7.5 mC. The combined resistance of the coil and the galvanometer is 0.50 Ω.Estimate the field strength of magnet.

Answer - 14 : -


Question - 15 : - Figure shows a metal rod PQ resting on the rails AB and positionedbetween the poles of a permanent magnet. The rails, the rod, and the magneticfield are in three mutual perpendicular directions. A galvanometer G connectsthe rails through a switch K. Length of the rod = 15 cm, B = 0.50 T, resistanceof the closed loop containing the rod = 9.0 mΩ. Assume the field to be uniform.
(a) Suppose K is open and the rod is moved with a speed of 12 cm s-1 inthe direction shown. Give the polarity and magnitude of the induced emf.
(b) Is there an excess charge built up at the ends of the rods when K is open ?What if K is closed ?
(c) With K open and the rod moving uniformly, there is no net force on theelectrons in the rod PQ even though they do experience magnetic force due tothe motion of the rod. Explain.
(d) What is the retarding force on the rod when K is closed ?
(e) How much power is required (by an external agent) to keep the rod moving atthe same speed (= 12 cm s_1) when K is closed ? How much power isrequired when K is open ?
(f) How much power is dissipated as heat in the closed circuit ? What is thesource of this power ?
(g) What is the induced emf in the moving rod if the magnetic field is parallelto the rails instead of being perpendicular ?

Answer - 15 : - (a) Using e = Bυl, we get
e = 0.5 x 12 x 10.2 x 15 x 102
= 9 x 10-3 V
The electrons in the rod will experience force along PQ, so end P becomespositive and Q becomes negative.
(b) On closing the key K,the number of electrons become more at end Q. Therefore, excess charge ismaintained by the continuous current.
(c) Magnetic force getscancelled by electric force due to excess charges of opposite sign at the endsof the rod.
(d) Retarding force, F =BIl =B eR l

Question - 16 : - An air-cored solenoid with length 30 cm, area of cross-section 25 cm2 andnumber of turns 500, carries a current of 2.5 A. The current is suddenlyswitched off in a brief time of 10-3 s. How much is the averageback emf induced across the ends of the open switch in the circuit ? Ignore thevariation in magnetic field near the ends of the solenoid.

Answer - 16 : -


Question - 17 : - (a) Obtain an expression for the mutual inductance between a longstraight wire and a square loop of side a as shown in Figure
(b) Now assume that the straight wire carries a current of 50 A and theloop is moved to the right with a constant velocity, υ = 10 m/s. Calculate theinduced e.m.f. in the loop at the instant when x = 0.2 m. Take a = 0.1 m andassume that the loop has a large resistance.

Answer - 17 : - (a) Consider a small portion of the coil of thickness dt at a distancet from the current carrying wire.Then the magnetic field strength experiencedby this portion

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