Question

An $\alpha$ particle is moving along a circle of radius $R$ with a constant angular velocity $\omega$. Point $\text{A}$ lies in the same plane at a distance $2R$ from the centre, and it records magnetic field produced by $\alpha$ particle. If the minimum time interval between two successive times at which $\text{A}$ records zero magnetic field is $t$, the angular velocity $\omega$ is:

• A. $\frac{2\pi }{t}$
• B. $\frac{2\pi }{3t}$
• C. $\frac{\pi }{3t}$
• D. $\frac{\pi }{t}$

Answer 1

The correct option is B $\frac{2\pi }{3t}$

We can assume $\alpha$ particle to be a current element in the circular path.

From Biot- Savart law,

$\overrightarrow{dB}=\frac{{\mu }_{0}i(\overrightarrow{dl}\times \overrightarrow{r})}{4\pi r^2}$

Here $\overrightarrow{dl}$ is in the direction of current i.e along the direction of velocity of $\alpha -$ particle at any position in the circular path.

The value of magnetic field will be zero when

$\overrightarrow{dl}\times \overrightarrow{r}=0$

Thus, the condition for $\overrightarrow{dl}\times \overrightarrow{r}=0$ is achieved when velocity of $\alpha -$ particle is parallel or antiparallel to the direction of position vector $\left(\overrightarrow{r}\right)$, as shown in figure at $\text{P}$ and $\text{Q}$.

At $\text{P}$ and $\text{Q}$ two tangents can be drawn from point $\text{A}$.

From geometry,

$\cos \theta =\frac{R}{2R}=\frac{1}{2}$

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

Since, arc $\text{PQ}$ subtends angle ${120}^{\circ }$ at centre, so time taken by $\alpha -$ particle to reach from $\text{P}$ to $\text{Q}$ is :

$t=\frac{\Delta \theta }{\omega }=\frac{\left(\frac{2\pi }{3}\right)}{\omega }$

$\therefore \omega =\frac{2\pi }{3t}$

Hence, option $\left(b\right)$ is correct.

Why this question? Tip: Focus on the direction of $\overrightarrow{dl}$ and then use Biot- Savart law to obtain the condition when magnetic field $\overrightarrow{dB}$ will become zero.