Question

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

• A. $\frac{2m{v}_0}{\mu (mg+q{E}_0)}$
• B. $\frac{m{v}_0}{\mu (mg+q{E}_0)}$
• C. $\frac{3m{v}_0}{\mu (mg+q{E}_0)}$
• D. $\frac{4m{v}_0}{\mu (mg+q{E}_0)}$

Answer 1

The correct option is B $\frac{m{v}_0}{\mu (mg+q{E}_0)}$

Electric force and weight $mg$ will be acting downwards,

$\therefore N=mg+q{E}_0.....\left(1\right)$

Magnetic force on the charge particle provides the required centripetal force,

$q{B}_{0}v=\frac{m{v}^2}{R}......\left(2\right)$

At some instant time $t$ , let velocity of the particle be $v$ and friction acts opposite to the direction of motion of the charge thus,

$-m\frac{dv}{dt}=\mu N......\left(3\right)$

From $\left(2\right)$ we get, $R=\frac{mv}{B_{0}q}...\left(4\right)$

From $\left(1\right)$ and $\left(3\right)$ we get,

$-m\frac{dv}{dt}=\mu (mg+q{E}_0)$

$-m{\int }_{v_0}^{0}dv=\mu (mg+q{E}_0){\int }_0^{t}dt$

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

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option (b) is the correct answer.