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 distance travelled by the particle before it comes to rest.

• A. $\frac{3m{v_0}^2}{2\mu (mg+q{E}_0)}$
• B. $\frac{2m{v_0}^2}{2\mu (mg+q{E}_0)}$
• C. $\frac{m{v_0}^2}{\mu (mg+q{E}_0)}$
• D. $\frac{m{v_0}^2}{2\mu (mg+q{E}_0)}$

Answer 1

The correct option is D $\frac{m{v_0}^2}{2\mu (mg+q{E}_0)}$

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

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

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,

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

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

we can write, $\frac{dv}{dt}=\frac{dv}{dl}\times \frac{dl}{dt}=v\frac{dv}{dl}$

$-mvdv=\mu (mg+q{E}_0)dl......\left(4\right)$

integrating both sides with proper limits,

${\int }_{v_0}^0-mvdv=\mu (mg+q{E}_0){\int }_0^{l}dl$

$l=\frac{m{v_0}^2}{2\mu (mg+q{E}_0)}$

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