A square loop of finite resistance at time $t = 0$ starts from rest moving parallel to x-axis with a constant acceleration $1 m / s^{2}$ as shown, in a space where magnetic field is changing with position and time according to relations.

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Initially current in loop is clockwise

Initially current in loop is anticlockwise

After $t >$ 1s, current in anticlockwise

At $t = 1 s$ , current in loop is zero

answer is A, C.

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Detailed Solution

Magnetic flux through the loop, $ϕ = B A (A = constant)$

$d B = \frac{\partial B}{\partial t} d t + \frac{\partial B}{\partial x} d x = \frac{\partial B}{\partial t} d t + \frac{\partial B}{\partial x} v d t = \frac{\partial B}{\partial t} d t + \frac{\partial B}{\partial x} a t d t$

$d B = \frac{\partial B}{\partial t} d t + \frac{\partial B}{\partial x} a t d t$

At $t = 0,$ $d B = \frac{\partial B}{\partial t} d t > 0$ . Inward magnetic flux linked with the loop increases. Hence by Lenz's law, direction of induced current is anticlockwise. Option (a) is correct, option (b) is wrong.

For zero induced current, change in magnetic flux linked with the loop over a small interval of time dt should be zero. It in turn means change in magnetic field, dB, should be zero.

$d B = d t (\frac{\partial B}{\partial t} + \frac{\partial B}{\partial x} a t) = 0$

$\Rightarrow t = \frac{(\frac{\partial B}{\partial t})}{- \frac{\partial B}{\partial x} a} = 1 s$ Option (c) is correct.

For $t > 1 s$ , $d B = (\frac{\partial B}{d t} d t + \frac{\partial B}{d x} a t d t) < 0$ . It implies inward magnetic flux linked with the loop decreases. By Lenz's law, induced current is clockwise. Option (d) is wrong.