Question

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

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

Answer 1

Answer: (B)

$\frac{2\pi }{3t}$

Magnetic field using Biot Savart law:
$\overrightarrow{B}=\frac{{\mu }_0}{4\pi }i\frac{(\overrightarrow{dl}\times \overrightarrow{r})}{{\overrightarrow{r}}^3}$

Though the expression above has current i, it can be recast into
$\overrightarrow{B}=\frac{{\mu }_0}{4\pi }q\frac{(\overrightarrow{v}\times \overrightarrow{r})}{{\overrightarrow{r}}^3}$
Point A shall record zero magnetic field (due to $\alpha$ particle) when
the -particle is at position P and Q as shown in figure since $\overrightarrow{v}\parallel \overrightarrow{r}$

For any other position of the $\alpha$ particle $\overrightarrow{v}$ is not parallel $\overrightarrow{r}$
$\therefore \overrightarrow{v}\times \overrightarrow{r}$ is not equal to zero, magnetic field is not equal to zero.
$\angle POQ={120}^0$
$\therefore \omega t={120}^0=\frac{2\pi }{3}$
$\Rightarrow \omega =\frac{2\pi }{3t}$