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 to the loop (directed into the paper). The loop is connected to a network of resistors each of value 3 ohms. The resistances of hte lead wire 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? Given the direction of current in the loop.

Answer

B, C, D

Explanation

The network behaves like a balanced wheatstone bridge.
The free electrons in the portion MN of the rod have a velocity v in the right direction. Applying Fleming's left hand rule, we find that the force on electron will be towards N. Hence, M will be +ve and N will be negative. Current will flow in clockwise direction.
the induced emf developed is given by
`e=vBl=vxx2xx0.1=0.2v`....(i)
Now, `e=IR`
`e=10^(-3)xx4xx4xx10^(-3)`amp....(ii)
from (i) and(ii)
`0.2v=4xx10^(-3)`
`:. v=(4xx10^(-3))/(0.2)=0.02m//s`.