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A rectangular loop of wire ABCD is kept close to an infinitely long wire carrying a current $ I (t) = I_0 (1 - \frac {t}{T}) for $ 0 \leq t \leq T$ and $ I (0) = 0$ for t > T, as shown in the fig. Find the total charge passing through a given point in the loop, in time T, The resistance of the loop is R.

(a) $\frac {\mu_0 L_2 I_2}{2\pi} \; log_e \; (\frac {L_1 + x}{x})$
(b) $ \frac {\mu_0 L_1 I_1}{2\pi} \; log_e \; (\frac {L_2 + x}{x})$
(c) $ \frac {\mu_0 L_1 I_1}{2\pi} \; log \; (\frac {L_2 + x}{x})$
(d) $ \frac {\mu_0 L_2 I_1}{2\pi} \; log_e \; (\frac {L_1 + x}{x})$

1 Answer

In the fig, XY is an infinitely long wire carrying current $ I (t) = \frac {I_0}{1 - \frac{t}{T}}$ for $ 0 \leq t \leq T$
If I(t) is current induced in the loop, then;
$ I(t) = \frac {d\phi / dt}{R} = \frac {dQ}{dt}$
Integrating on both sides;
$ Q (L_1) - Q (t_2) = \frac {1}{R} [ \phi (t_1) - \phi (t_2)]$
$ \phi (T_1) = L_1 \frac {\mu_0}{2\pi} \; \int\limits_x^{L_2 + x} \; \frac {dx^1}{x^1} \; I (t_1)$
$ = \frac {\mu_0 L_1}{2\pi}\; I (t_1) \; log_e \; (\frac {L_2 + x}{x})$
The magnitude of charge would be;
$ Q = \frac {\mu_0 L_1}{2\pi} \; log_e \; (\frac {L_2 + x}{x}) [I_1 - 0]$
$ = \frac {\mu_0 L_1 I_1}{2\pi} \; log_e \; (\frac {L_2 + x}{x})$
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