Question

A magnetic field $\overrightarrow{B}=B_0\hat{j}$ exists in the region $a<x<2a$ and $\overrightarrow{B}=-B_0\hat{j}$ in the region $2a<x<3a$, where $B_0$ is a positive constant. A positive point charge moving with a velocity $\overrightarrow{v}=v_0\hat{i}$ where $v_0$ is a positive constant, enters the magnetic field at $x=a$. The trajectory of the charge in this region can be like,

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Magnetic force on the moving charge in the magnetic field is given by

$\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$

From $a$ to $2a$:

$\overrightarrow{B}=B_0\hat{j}\text{and}\overrightarrow{v}=v_0\hat{i}$

$\therefore \overrightarrow{F}=q{v}_0{B}_0(\hat{i}\times \hat{j})=q{v}_0{B}_0\hat{k}$

Thus, the particle gets deflected towards $+z-\text{axis}$, which is vertically upward.

From $2a$ to $3a$:

$\overrightarrow{B}=-B_0\hat{j}\text{and}\overrightarrow{v}=v_0\hat{i}$

$\therefore \overrightarrow{F}=q{v}_0{B}_0[\hat{i}\times (-\hat{j}\left)\right]=q{v}_0{B}_0(-\hat{k})$

So, in the region $2a<x<3a$, the force on the charge will be in $-z$ direction, which is vertically downward.

For $a<x<2a$, path will be concave upward because of upward force and for $2a<x<3a$, path will be concave downward.

Moreover, there must not be any kink in the path at $x=2a$. Combining all we get the graph as


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