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A square loop of side 20 cm and resistan...

A square loop of side 20 cm and resistance `1 Omega` is moved towards right with a constant speed `v_0`. The right arm of the loop is in a uniform magnetic field of 5 T. The field is perpendicular to the plane of the loop and is going into it. The loop is connected to a network of resistors each of value `4 Omega.` What should be the value of `v_0` so that a steady current of 2 mA flows in the loop ?

A

`10^(-2) cm//s`

B

`1 cm//s`

C

`1 m//s`

D

`10^2 m//s`

Text Solution

Verified by Experts

The correct Answer is:
B

KVL
`+ (2xx10^(-3)) xx1 -v_0 + 8 xx10^(-3)=0`
`implies v_0 =10 xx10^(-3)m//s`
Bvl = `v_0`
`v=(v_0)/(Bl)=(10xx10^(-3))/(5xx20 xx10^(-2))=10 xx10^(-3)` = 1 cm /s
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