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### A long straight wire carries a current, I = 2 ampere. A semi-circular ....

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) Maximum current through R is $1.2 \times 10^{-6}$ ampere
(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.

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

### A man starts walking from the point P (-3, 4) ....

A man starts walking from the point P (-3, 4), touches the x-axis at R, and then turns to reach at the point Q (0, 2). The man is walking at a constant speed. If the man reaches the point Q in the minimum time, then $50 [(PR)^2 + (RQ)^2 ]$ is equal to _ _ _ _ . Solution For time to be minimum at constant speed, the directions must be symmetric. In other words, the angles made by PR and RQ with the vertical must be the same just like in the law of reflection in optics. $tan \theta = \frac {MP}{MR} = \frac {NQ}{NR}$ $\Rightarrow \frac {3-r}{4} = \frac {r}{2}$ $\Rightarrow r=1$ So, $R \equiv ( - 1,0)$ Now, $50(PR^2+RQ^2)=50[(4+16)+(1+4)]=1250$