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