Question

A helium nucleus makes a full rotation in a circle of radius $0.8\text{m}$ in $2\text{s}$. The value of magnetic field at the centre of the circle is

• A. $\left(\frac{{10}^{-19}}{{\mu }_0}\right)\text{T}$
• B. $\left({10}^{-19}{\mu }_0\right)\text{T}$
• C. $(2\times {10}^{-10}{\mu }_0)\text{T}$
• D. $\left(\frac{2\times {10}^{-10}}{{\mu }_0}\right)\text{T}$

Answer 1

The correct option is B $\left({10}^{-19}{\mu }_0\right)\text{T}$

Helium nucleus completes the rotation in just $2\text{s}$. So through any cross-section of the wire in $2\text{s}$, one helium nuclei will pass.

So the charge flowing in $2\text{s},q=2\times e$

$q=2\times 1.6\times {10}^{-19}\text{C}$

$\therefore \text{current,}I=\frac{q}{t}=\frac{2\times 1.6\times {10}^{-19}}{2}$

$I=1.6\times {10}^{-19}\text{A}$

So, the magnetic field at the centre of the circle is given by

$B=\frac{{\mu }_{0}I}{2R}=\frac{{\mu }_0\times 1.6\times {10}^{-19}}{2\times 0.8}$

$B={\mu }_0\times {10}^{-19}\text{T}$

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

Why this question? If a $q$ amount of charge is moving in a circle and completes rotation during one time period then current in the circuit is $i=\frac{q}{T}=qf$ $f\to$ frequency, $f=\frac{1}{T}$