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A field of strenght 5 xx 10^(4)//pi ampe...

A field of strenght `5 xx 10^(4)//pi` ampere turns`//`meter acts at right angles to the coil of `50` turns of area`10^(-2) m^(2)`. The coil is removed from the field in `0.1` second. Then the induced `e.m.f` in the coil is

A

0.1 V

B

0.2 V

C

1.96 V

D

0.98V

Text Solution

Verified by Experts

The correct Answer is:
A
`B=mu_0H=(mu_0xx5xx10^6)/pi`
`e=(NBA)/(time)=(50xx2xx10^-2xx10^-2)=-.11`
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NARAYNA-ELECTROMAGNETIC INDUCTION-EXERCISE-1 (C.W)
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