question_answer

A square metal wire loop of side 10 cm and resistance 1 ohm is moved with a constant velocity \[{{v}_{0}}\] in a uniform magnetic field of induction \[B=2\,weber/{{m}^{2}}\] as shown in the figure. The magnetic field lines are perpendicular to the plane of the loop (directed into the paper). The loop is connected to a network of resistors each of value 3 ohms. The resistances of the lead wires OS and PQ are negligible. What should be the speed of the loop so as to have a steady current of 1 milliampere in the loop? Give the direction of current in the loop.      

A) 2.3 m/s

B) 1.2 m/s

C) 0.3 m/s

D) 0.02 m/s

Correct Answer: D

Solution :

[d] The network behaves like a balanced wheatstone bridge. The induced emf developed is given by \[e=vB\ell =v\times 2\times 0.1=0.2v\]             ...(i) Now,  \[e=IR\] \[e={{10}^{-3}}\times 4=4\times {{10}^{-3}}\,\,\,\,amp\]          ...(ii) From (i) and (ii), \[0.2\,v=4\times {{10}^{-3}}\]   \[\therefore \,\,\,\,\,v=\frac{4\times {{10}^{-3}}}{0.2}=0.02m/s\]