Question

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

• A. 2 s
• B. 3 s
• C. 5 s
• D. 6 s

Answer 1

Answer: (C)

Given inclination of plane,

Mass of block, $m=20 \,g=0.02 \,kg$
charge on the block, $q=4 \,mC =4 \times 10^{-3} C$
Magnetic field, $B=1 \,T$
Magnetic force on the charge particles, $F=B q v$
Particle will leave the inclined plane, when
$F=mg \,\cos\,\theta \Rightarrow B q v=mg \,\cos \,\theta \Rightarrow v=\frac{m g \cos \theta}{q B}$
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{m g \cos \theta}{q B \cdot g \sin \theta}=\frac{m \cot \theta}{q B} $
$t=\frac{0.02 \cot 45^{\circ}}{4 \times 10^{-3} \times 1}=5 s \left[\because \cot 45^{\circ}=1\right]$