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A conducting bar is pulled with a consta...

A conducting bar is pulled with a constant speed `v` on a smooth conducting rail. The region has a steady magnetic field of induction `B` as shown in the figure. If the speed of the bar is doubled then the rate of heat dissipation will

A

remain constant

B

become quarter of the initial value

C

become four fold

D

get doubled

Text Solution

Verified by Experts

The induced emf between `A` and `B=E=Blv`
`rArr ` The induced current `=i=(E)/(R )`
`rArr i=(B l v)/(R )`
The electrical power `=P=i^(2)R=(B^(2)l^(2)v^(2))/(R )`
Since `v` is doubled, the electrical power, becomes four times. Since heat dissipation per second is proportional to electrical power, it becomes four fold.
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