question_answer

A rectangular loop has a sliding connector PQ of length l and resistance \[R\Omega \], and it is moving with a speed v as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents \[{{I}_{1}},{{I}_{2}}\]and I are

A) \[{{I}_{1}}={{I}_{2}}=\frac{Blv}{6R},I=\frac{Blv}{3R}\]

B) \[{{I}_{1}}=-{{I}_{2}}=\frac{Blv}{R},I=\frac{2Blv}{R}\]

C) \[{{I}_{1}}={{I}_{2}}=\frac{Blv}{3R},I=\frac{2Blv}{3R}\]

D) \[{{I}_{1}}={{I}_{2}}=I=\frac{Blv}{R}\]

Correct Answer: C

Solution :

[c]: Emf induced across PQ is \[\varepsilon =Blv\].

The equivalent circuit diagram is as shown in the figure.

Applying Kirchhoff?s first law at junction Q, we get

\[I={{I}_{1}}+{{I}_{2}}\]                                            ...(i)

Applying Kirchhoff?s second law for the closed loop PLMQP, we get

\[-{{I}_{1}}R-IR+\varepsilon =0\]

\[{{I}_{1}}R+IR=Blv\]                                      ...(ii)

Again, applying Kirchhoff?s second law for the closed loop PONQP, we get

\[-{{I}_{2}}R-IR+\varepsilon =0\]

\[{{I}_{2}}R+IR=Blv\]                                      ...(iii)

Adding equations (ii) and (iii), we get

\[2IR+{{I}_{1}}R+{{I}_{2}}R=2Blv\]

\[2IR+R({{I}_{1}}+{{I}_{2}})=2Blv\]

\[2IR+IR=2Blv\]                            (Using (i))

\[3IR=2Blv\]

\[I=\frac{2Blv}{3R}\]                                            ...(iv)

Substituting this value of I in equation (ii), we get

\[{{I}_{1}}=\frac{Blv}{3R}\]

Substituting the value of I in equation (iii), we get

\[{{I}_{2}}=\frac{Blv}{3R}\]

Hence, \[{{I}_{1}}={{I}_{2}}=\frac{Blv}{3R},I=\frac{2Blv}{3R}\]