Question

A particle of charge $q$ and mass $m$ moving under the influence of uniform electric field $E\hat{i}$ and a uniform magnetic field $-B\hat{i}$ moves from point $\text{P}(0,a)$ to $\text{Q}(2a,a)$ where $\text{O}(0,0)$ is origin. The velocity of the particle at Point $\text{P}$ is $v\hat{i}$ and at $\text{Q}$ is $2v\hat{i}$. The rate of work done by the electric field & magnetic field on particle at point $\text{P}$ will be respectively:

• A. $\frac{3}{4}\frac{m{v}^3}{a},0$
• B. $\frac{3}{4}\frac{m{v}^3}{a},\frac{3}{2}\frac{m{v}^3}{a}$
• C. $\frac{3}{2a}m{v}^3,0$
• D. $0,\frac{m{v}^3}{2a}$

Answer 1

The correct option is A $\frac{3}{4}\frac{m{v}^3}{a},0$

The motion of particle will be a straight line because the velocity of particle is parallel to direction of $\overrightarrow{E}$ and antiparallel to direction of $\overrightarrow{B}$. The magnetic force acting on particle will be,
$\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})=0$

Thus workdone by magnetic force will be zero throughout motion of charge.


Increase in kinetic energy of particle,

$\Delta KE=\frac{1}{2}m(2v{)}^2-\frac{1}{2}m(v^2)$

$\Delta KE=\frac{3}{2}m{v}^2$

Workdone by $\overrightarrow{E}$ in going from $\text{P}$ to $\text{Q}$ is,

$W_e=F_e\times d$

$W_e=qE\times \left(2a\right)=2qEa$

Using Work-energy theorem,

$W_e+W_m=\Delta KE$

or, $2qEa+0=\frac{3}{2}m{v}^2$

$E=\frac{3m{v}^2}{4qa}$

Now the rate of work done is power delivered and it will be only due to electric field, because the power delivered by magnetic field is zero.

$\Rightarrow P_e=\overrightarrow{F_e}.\overrightarrow{v}$

$P_e=\left(qE\hat{i}\right).\left(v\hat{i}\right)=qEV$

$P_e=q\left(\frac{3v^{2}m}{4qa}\right)v=\frac{3m{v}^3}{4a}$

$P_m=0$