A conducting rod, with a resistor of resistance R. is pulled with constant speed v on a smooth conducting rail as shown in figure. A constant magnetic field `vecB` is directed into the page. If the speed of the bar is doubled, by what factor does the rate of heat dissipation across the resistance R change?

Options

A. 0
B. `sqrt(2)`
C. `2`
D. `4`

Answer

D

Explanation

`Emf` = VBL
` I=(VBL)/(R )`
heat `=I^2 R = (V^2 B^2 L^2)/( R)`
given ` V^1 = 2V `
So ` (H^1)/(H ) =4`