Question

A positively charged particle of mass m and charge q is projected on a rough horizontal x-y plane surface with z-axis in the vertically upward direction. Both electric and magnetic fields are acting in the region and given by $\overrightarrow{E}=-E_0\hat{k}$ and $\overrightarrow{B}=-B_0\hat{k}$, respectively. The particle enters into the field at $(a_0,0,0)$ with velocity $\overrightarrow{v}=v_0\hat{j}$. The particle starts moving in some curved path on the plane. The coefficient of friction between the particle and the plane is given as $\mu$. If the time when the particle will come to rest is $t=\frac{m{v}_0}{x\mu (mg+q{E}_0)}$. Find x.

Answer 1

Force of electric field will act in downward direction, so normal force between particle and surface will be as:
$N=mg+q{E}_0$ .... (i)

Let at any time velocity of the particle is v, then
$-m\frac{dv}{dt}=\mu N$ .... (ii)

From Eqs (i) and (ii),
$-m\frac{dv}{dt}=\mu (mg+q{E}_0)$
$-m{\int }_{v_0}^{0}dv=\mu (mg+q{E}_0){\int }_0^{l}dt$

Thus, $t=\frac{m{v}_0}{\mu (mg+q{E}_0)}$