Question

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

Answer 1

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.