A long straight wire carries a current, I = 2 ampere. A semi-circular conducting rod is placed beside it on two conducting parallel rails of negligible resistance. Both the rails are parallel to the wire. The wire, the rod and the rails lie in the same horizontal plane, as shown in the figure. Two ends of the semi-circular rod are at distances 1 cm and 4 cm from the wire. At time t = 0, the rod starts moving on the rails with a speed v = 3.0 m/s (see the figure).

A resistor $R = 1.4 \Omega$ and a capacitor $C_0 = 5.0 \mu F$ are connected in series between the rails. At time t = 0, $C_0$ is uncharged. Which of the following statement(s) is(are) correct? [$\mu_0 = 4\pi \times 10^{-7}$ SI units. Take ln 2 = 0.7]

A resistor $R = 1.4 \Omega$ and a capacitor $C_0 = 5.0 \mu F$ are connected in series between the rails. At time t = 0, $C_0$ is uncharged. Which of the following statement(s) is(are) correct? [$\mu_0 = 4\pi \times 10^{-7}$ SI units. Take ln 2 = 0.7]

(B) Maximum current through R is $3.8 \times 10^{-6}$ ampere

(C) Maximum charge on capacitor $C_0$ is $8.4 \times 10^{-12}$ coulomb

(D) Maximum charge on capacitor $C_0$ is $2.4 \times 10^{-12}$ coulomb

*Solution*

The semicircular arc can be replaced by a straight wire joining end to end. Taking a small element of length dx in this straight wire at a distance x from the I current carrying wire.

Induced e.m.f. $dE = \frac{{{\mu _0}}}{{4\pi }}\frac{{2I}}{x}.dx.v$

$E = \int\limits_{1 \times {{10}^{ - 2}}}^{4 \times {{10}^{ - 2}}} {\frac{{{\mu _0}}}{{4\pi }}2Iv\frac{{dx}}{x}} = \frac{{{\mu _0}}}{{4\pi }}2Iv\left. {\ln x} \right|_{1 \times {{10}^{ - 2}}}^{4 \times {{10}^{ - 2}}}$

$ \Rightarrow E = {10^{ - 7}} \times 2 \times 2 \times 3 \times \ln 4 = 1.68 \times {10^{ - 6}}V$

Current is maximum in the RC circuit at t = 0 when the capacitor is short circuit.

${i_{\max }} = \frac{E}{R} = \frac{{1.68 \times {{10}^{ - 6}}}}{{1.4}} = 1.2 \times {10^{ - 6}}A$

Charge is maximum in the RC circuit at steady state when the capacitor is open circuit.

${q_{\max }} = {C_0}E = 5.0 \times {10^{ - 6}} \times 1.68 \times {10^{ - 6}} = 8.4 \times {10^{ - 12}}C$