SSC JE Electrical Engineering 2013 Paper

Question : 116 of 200

Marks: +1, -0

The total flux at the end of a long permanent bar magnet is 100 × 10

^{–6}

Wb. The end of this magnet is withdrawn through a 1000 turn coil in

1/20

seconds. The induced e.m.f. in the coil is

Solution:

induced Emf

={ϕ.N}/t

={100×10^{-6}×100}/{1/20}

=2.0V

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