Question

In gravity free space, a bead of charge $1\,\mu C$ and mass $3\,mg$ is threaded on a rough rod of friction coefficient $\mu =0.3.$ A magnetic field of magnitude $0.2\,T$ exists perpendicular to the rod. The bead is projected along the rod with a speed of $4\,m/s$ . How much distance (in $m$ ) will the bead cover before coming to rest?

Answer 1

$N=qvB$
$-\mu qvB=\frac{m d v}{d t}=mv\frac{d v}{d S}$
$\mu qBS=mv$
$S=\frac{3 \times 10^{- 6} \times 4}{0 . 3 \times 10^{- 6} \times 0 . 2}=200\,m$