A) 1 cm/sec
B) 2 cm/sec
C) 3 cm/sec
D) 4 cm/sec
Correct Answer: B
Solution :
Equivalent resistance of the given Wheatstone bridge circuit (balanced) is 3W so total resistance in circuit is \[R=3+1=4\Omega \]. The emf induced in the loop \[e=Bvl\]. So induced current \[i=\frac{e}{R}=\frac{Bvl}{R}\,\,\,\] \[\Rightarrow \,\,\,{{10}^{-3}}=\frac{2\times v\times (10\times {{10}^{-2}})}{4}\Rightarrow v=2cm/sec.\]You need to login to perform this action.
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