Question

A block of mass $m$ and charge $q$ is released on a long smooth inclined plane. Magnetic field $B$ is constant, uniform and out of the plane of the paper as shown. Find the time from the start when block leaves contact with the surface.

• A. $\frac{m \cos \theta }{q B}$
• B. $\frac{m \cos e c \theta }{q B}$
• C. $\frac{m \cot \theta }{q B}$
• D. $\frac{m \tan \theta }{q B}$

Answer 1

Answer: (C)

Block will lose contact with surface when force due to magnetic field will become equal to $mg \cos\theta $
$qvB=mg \cos\theta $
$v=\frac{mg \cos\theta }{qB}=u+at$ ( along inclined plane )
$v=\frac{mg \cos\theta }{qB}=\left(\right.g\text{ sin}\theta \left.\right)t$
$t= \, \frac{\text{m cot} \theta }{qB}$