Question

A small block of mass $20g$ and charge $4mC$ is released on a long smooth inclined plane of inclination angle of $45^{\circ }$, A uniform horizontal magnetic field of $1T$ 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

• A. 2 s
• B. 3 s
• C. 5 s
• D. 6 s

Answer 1

Answer: (C)

Given inclination of plane,

Mass of block, $m=20g=0.02kg$
charge on the block, $q=4mC=4\times 10^{-3}C$
Magnetic field, $B=1T$
Magnetic force on the charge particles, $F=Bqv$
Particle will leave the inclined plane, when
$F=mg\cos \theta \Rightarrow Bqv=mg\cos \theta \Rightarrow v=\frac{mg\cos \theta }{qB}$
Time taken to reached at the velocity $v$ is given by
$v=0+g\sin \theta t[\because u=0,a=g\sin \theta ]$
$t=\frac{v}{g\sin \theta }=\frac{mg\cos \theta }{qB\cdot g\sin \theta }=\frac{m\cot \theta }{qB}$

$t=\frac{0.02\cot 45^{\circ }}{4\times 10^{-3}\times 1}=5s\left[\because \cot 45^{\circ }=1\right]$