1.

A rod of 5 cm length is moving perpendicular to uniform magnetic field of intensity 2 xx 10^(-4) "Wbm"^(-2). If the acceleration of rod is 2 ms^(-2), the rate of increasing of the induced emf is .......

Answer»

`20xx10^(-4) Vs^(-2)`
`20xx10^(-4)V`
`20xx10^(-4)` Vs
`2 xx10^(-5)Vs^(-1)`

SOLUTION :The induced EMF `|epsilon|=Bvl`
`therefore (d|epsilon|)/(DT)=d/(dt)(Bvl)`
`=B(dv)/(dt)l` [ `because` B and l are constant]
=Bal[`because (dv)/(dt)`=a=acceleration]
`=2XX10^(-2)xx2xx5xx10^(-2)`
`=20xx10^(-4)Vs^(-1)`


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