A small block of mass 20 g and charge 4 mc is released on a long smooth inclined plane of inclination angle of `45^(@)` A uniform horizontal magnetic field of 1 T is acting parallel to the surface as shown in the figure . The time from the start when the block loses contact with the surface of the plane is

Options

A. 2 s
B. 3 s
C. 5 s
D. 6 s

Answer

C

Explanation

Given inclination of plane

Mass of block m =20 g =0.02 kg
charge on the block q=4 mc` = 4 xx 10^(-3) c` Magnetic force on the charge particles F = Bqv
Particle will leave inclined plane when ,
`F= mg cos theta rArr Bav = mg cos theta rArr v= (mg cos theta )/( qB)`
Time taken to reached at the velocity v is given by
`v=0 + g sin theta t `
`t = (v )/( g sin theta ) = (mg cos theta )/( qB . g sin theta) = (m cos theta )/( qB ) `
`t = (0.02 cot 45^(@))/( 4 xx 10^(-3) xx l ) = 5 s`