A circuit ABCD is held perpendicular to the uniform magnetic field of B=5×10-2Textending over the region PQRS and directed into the plane of the paper. The circuit is moving out of the field at a uniform speed of 0.2 ms-1 for 1.5 s. During this time, the current in the 5Ω resistor is

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Q.

A circuit ABCD is held perpendicular to the uniform magnetic field of $B = 5 \times 10^{- 2} T$ extending over the region PQRS and directed into the plane of the paper. The circuit is moving out of the field at a uniform speed of $0.2 m s^{- 1}$ for 1.5 s. During this time, the current in the $5 Ω$ resistor is

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a

0.9 mA from C to B

b

0.9 mA from B to C

c

`0.6 mA from C to B

d

0.6 mA from B to C

answer is A.

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Detailed Solution

$I = \frac{B l v}{R}$

$I = \frac{5 \times 10^{- 2} \times 0.3 \times 0.2}{5} A = 0.6 m A$

Area and flux are decreasing. So, current flows to increase the flux. Clearly, current should be clockwise. So, it flows from .B
to C through $5 Ω$

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A circuit ABCD is held perpendicular to the uniform magnetic field of B=5×10-2Textending over the region PQRS and directed into the plane of the paper. The circuit is moving out of the field at a uniform speed of 0.2 ms-1 for 1.5 s. During this time, the current in the 5Ω resistor is