Question

When an electron of mass $m$ and charge $-e$ is released from rest in a room, it starts with initial acceleration $a_0$ towards West. When it is projected towards North with a speed $3v_0$ it moves with an initial acceleration $2a_0$ towards West. The electric and magnetic fields in the room are -

• A. $\frac{2m{a}_0}{e}$ East ; $\frac{m{a}_0}{3e{v}_0}$ Upward
• B. $\frac{m{a}_0}{e}$ East ; $\frac{2m{a}_0}{e{v}_0}$ Upward
• C. $\frac{m{a}_0}{e}$ East ; $\frac{3m{a}_0}{e{v}_0}$ Upward
• D. $\frac{m{a}_0}{e}$ East ; $\frac{m{a}_0}{3e{v}_0}$ Upward

Answer 1

The correct option is D $\frac{m{a}_0}{e}$ East ; $\frac{m{a}_0}{3e{v}_0}$ Upward

In the first case, when the electron is released from rest:

There will be only electric force acting on the electron as $v=0$, so, magnetic force,

${\overrightarrow{F}}_m=-e(\overrightarrow{v}\times \overrightarrow{B})=0$

Also, electric force, $F_e=-eE$

$\therefore a=\frac{F_e}{m}=\frac{-eE}{m}=a_0\text{(Toward west)}$

$\Rightarrow E=\frac{m{a}_0}{e}\left(\text{Towards east}\right)$

In the second case, when the electron is released with a speed of $3v_o$ towards North:



As the net acceleration of $2a_0$ is toward West, the electric force and the magnetic force acts in the West.

So, acceleration due to electric force will be same as in the first case.

Hence, acceleration due to magnetic force will be $(2a_0-a_0)=a_0$.

$\therefore e\left(3v_o\right)B=m{a}_0$

$\Rightarrow B=\frac{m{a}_0}{3e{v}_0}$

From right-hand thumb rule, the direction of $B$ will be upward.