A metal rod `1/sqrtpi `m long rotates about one of its ends perpendicular to a plane whose magnetic induction is 4 x 10^{-3} T. Calculate the number of revolutions made by the rod per second if the e.m.f. induced between the ends of the rod is 16 mV.

Advertisement Remove all ads

#### Solution

`L=1/sqrtpim="radius"`

B = 4x10^{-3}T

e = 16mv

=16x10^{-3}v

Area covered in one revolution

A=πL^{2}

`=pi(1/sqrtpi)^2`

=1m^{2}

e=BfA

`:.f=e/(BA)`

`=(16xx10^(-3))/(4xx10^(-3)xx1)`

f=4Hz

Concept: Electromagnetic Induction

Is there an error in this question or solution?

#### APPEARS IN

Advertisement Remove all ads