Question

A uniform magnetic field of magnitude $1\text{T}$ exists in the region $y\ge 0$ along $+vez-$ direction $\left(\hat{k}\right)$ as shown in figure. A charged particle of $1\text{C}$ is projected from the point $(-\sqrt{3},-1)\text{m}$ towards the origin with speed $1\text{m/s}$. If mass of particle is $1\text{kg}$, then find the coordinates of the centre of circle in which particle will move.

• A. $(1,\sqrt{3})\text{m}$
• B. $(1,-\sqrt{3})\text{m}$
• C. $(\frac{1}{2},-\frac{\sqrt{3}}{2})\text{m}$
• D. $(\frac{\sqrt{3}}{2},\frac{-1}{2})\text{m}$

Answer 1

The correct option is C $(\frac{1}{2},-\frac{\sqrt{3}}{2})\text{m}$

Given:

$B=1\text{T};q=1\text{C};m=1\text{kg};v=1\text{m/s}$

Using the equation of magnetic force,

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

From the right- hand curl rule, the direction of magnetic force $F$ acting on the charged particle will be as shown in the diagram.


From the geometry of initial position of the particle,

$\tan \theta =\left|\frac{-\sqrt{3}}{-1}\right|=\sqrt{3}$

$\therefore \theta ={60}^{\circ }$

Thus, $\overrightarrow{F}$ makes angle ${30}^{\circ }$ from $y-$ axis as shown in figure.

Here, $\text{OC}$ is the radius $r$ of circular path because $\overrightarrow{F}\perp \overrightarrow{v}$, so,

$r=\frac{mv}{qB}=\frac{1\times 1}{1\times 1}=1\text{m}$

Thus, coordinates of centre will be,

$C(x,y)=(+r\sin {30}^{\circ },-r\cos {30}^{\circ })$

$\Rightarrow C(x,y)=(1\sin {30}^{\circ },-1\cos {30}^{\circ })=(\frac{1}{2},-\frac{\sqrt{3}}{2})\text{m}$

Therefore, option $\left(C\right)$ is the right answer.