Question

A particle of specific charge $\alpha$ starts moving from the origin under the action of an electric field $\overrightarrow{E}=E_0\hat{i}$ and magnetic field $\overrightarrow{B}=B_0\hat{k}$. Its velocity at $(x_0,y_0,0)$ is $(4\hat{i}+3\hat{j})$. The value of $x_0$ is:

• A. $\frac{5\alpha }{2B_0}$
• B. $\frac{13\alpha E_0}{2B_0}$
• C. $\frac{16\alpha E_0}{E_0}$
• D. $\frac{25}{2\alpha B_0}$

Answer 1

The correct option is D $\frac{25}{2\alpha B_0}$

Let $m\&q$ be the mass and charge of the particle respectively.

We know that work done by magnetic field remains zero.

$\therefore W_B=0$.

And, $W_e=\overrightarrow{F_e}\cdot \overrightarrow{d}=q{E}_0{x}_0$

$\Delta KE=\frac{1}{2}m{v}^2-0=\frac{1}{2}m{v}^2$

From work- energy theorem,

$W_E+W_B=\Delta KE......\left(1\right)$

$W_E+0=\Delta KE$

$E_{0}q{x}_0=\frac{1}{2}m{v}^2$

$x_0=\frac{v^2}{2\left(\frac{q}{m}\right)E_0}=\frac{v^2}{2\alpha E_0}$

Here, $v=\sqrt{v_x^2+v_y^2}=\sqrt{4^2+3^2}=5\text{m/s}$
$\alpha =q/m$

$\therefore x_0=\frac{5^2}{2\alpha E_0}=\frac{25}{2\alpha E_0}$

Hence, option $\left(d\right)$ is the correct answer.