Two straight infinitely long current wires lying along the x-axis and the y-axis carry currents 100A and 200A as indicated in the adjoining figure. A rigid square metallic wire frame ABCD of side 10cm and resistance (R = 10 $Ω$ ) is connected to a capacitor C1 = 1mF as shown in the figure. The wire frame, which lies in the x-y plane, moves with a constant velocity 30 $(\hat{i} + \hat{j}) m / s$ . At the instant, when the point A is located at (20cm, 20cm) the charge $q_{0}$ on the capacitor (as shown in the figure) is 0.2 $μ$ C. At the above mentioned instant,

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The current flowing in the circuit is given by $10^{- 4}$ amp.

The net emf is anticlockwise and is given by $10^{- 5} V$ .

The current flowing in the circuit is given by $10^{- 5}$ amp.

The net emf is anticlockwise and is given by $10^{- 4} V$ .

answer is A, B.

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Detailed Solution

The emf induced in the anticlockwise direction in the loop is
$\frac{μ_{0} 100}{2 π .2 \times 0.1} \times 0.1 \times 30 - \frac{μ_{0} 100}{2 π \times 3 \times 0.1} \times 0.1 \times 30$
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$= \frac{μ_{0}}{2 π} \times 100 \times 30 (\frac{1}{2} - \frac{1}{3}) = 10^{- 4} V$
The net emf is anticlockwise and is given by $10^{- 4} V$ .
Apply kirchoff law along the loop.
$\begin{array}{l} + \frac{q}{c} + i R = 10^{- 4} \\ i = [10^{- 4} - \frac{q}{c}] \frac{1}{R} = 10^{- 5} A \end{array}$