Question

A block of mass $m$ and charge $q$ is released on a long, smooth inclined surface. The magnetic field $B$ is constant, uniform, horizontal and parallel to the surface as shown. If the time from start when the block loses contact with the surface is $\frac{m\cot \theta }{k\cdot qB}$, then find $k$.

Answer 1

Answer: 1

$f_B=mg\cos \theta$
$qvB=mg\cos \theta$
$q[g(\sin \theta t)]B=mg\cos \theta$
$\Rightarrow t=\frac{m\cot \theta }{qB}$