Question

When a proton is released from rest in a room, it starts with an initial acceleration $a_0$ towards west. When it is projected towards north with a speed $v_0$ it moves with an initial acceleration $3a_0$ towards west. The electric and magnetic fields in the room are:

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

Answer 1

The correct option is C $\frac{m{a}_0}{e}\text{West};\frac{2m{a}_0}{e{v}_0}\text{Down}$

Initially, when it is released from rest,

$\Rightarrow \overrightarrow{E}=\frac{\overrightarrow{F}}{q}=\frac{m{a}_0}{e}\text{(west)}$

The same electrostatic force will act on it in the second case.

In the second case, the initial acceleration is directed towards west and so will the net force $F_{net}=3m{a}_0$ acting on the proton.

Now, the magnetic force in the second case will be,

$\Rightarrow \left|\overrightarrow{F_B}\right|=3m{a}_0-m{a}_0=2m{a}_0\text{(west)}$



According to Fleming's left hand rule, $\overrightarrow{B}$ is directed downwards.

Also, $\overrightarrow{F_B}=q(\overrightarrow{v}\times \overrightarrow{B})\text{(west)}$

$\Rightarrow ev{B}_0=2m{a}_0$

Therefore, the magnetic field will be,

$\overrightarrow{B}=\frac{2m{a}_0}{e{v}_0}\text{(down)}$

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