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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 isA. `0.1 V`B. `0.2 V`C. `1.96 V`D. `0.98 V`

Answer» Correct Answer - A
`B = mu_(0) H = (mu_(0) xx 5 xx 10^(4))/(pi)`
`e = (NBA)/("time") = (50 xx 2 xx 10^(-2) xx 10^(-2))/(0.1) = 0.11`


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