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 -                            [AIEEE 2010]

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

B) \[{{I}_{1}}=-{{I}_{2}}=\frac{B/v}{R},I=\frac{2B/v}{3R}\]

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

D) \[{{I}_{1}}={{I}_{2}}=I=\frac{B/v}{R}\]

Correct Answer: C

Solution :

[c]

\[I=\frac{2BVl}{3R}\]            \[{{I}_{1}}={{I}_{2}}=\frac{BVl}{3R}\]