JEE Physics 2023 Question with Solution

Given: An electron revolves with speed $6.76 \times 10^{6}\ m s^{-1}$ in an orbit of radius $0.52\ \AA$.

Find: The magnetic field at the nucleus.

A revolving electron constitutes a current loop. The magnetic field at the centre of a circular current loop is

$$B = \frac{\mu_0 I}{2r}$$

Current is charge per time period:

$$I = \frac{e}{T}$$

Time period of revolution is

$$T = \frac{2\pi r}{v}$$

So,

$$I = \frac{ev}{2\pi r}$$

Substituting in the magnetic field formula,

$$B = \frac{\mu_0}{2r} \cdot \frac{ev}{2\pi r} = \frac{\mu_0 e v}{4\pi r^2}$$

Now substitute the numerical values:

$$\mu_0 = 4\pi \times 10^{-7}\ H m^{-1}, \quad e = 1.6 \times 10^{-19}\ C$$ $$v = 6.76 \times 10^{6}\ m s^{-1}, \quad r = 0.52\ \AA = 0.52 \times 10^{-10}\ m$$

Therefore,

$$B = \frac{4\pi \times 10^{-7} \times 1.6 \times 10^{-19} \times 6.76 \times 10^{6}}{4\pi \times (0.52 \times 10^{-10})^2}$$

Cancelling $4\pi$,

$$B = \frac{10^{-7} \times 1.6 \times 6.76 \times 10^{-13}}{(0.52)^2 \times 10^{-20}}$$ $$B = \frac{10.816 \times 10^{-20}}{0.2704 \times 10^{-20}} \approx 40\ T$$

Therefore, the magnetic field produced at the nucleus is $40\ T$.

Given: A revolving electron of speed $v$ moves in a circular orbit of radius $r$.

Find: The magnetic field at the nucleus.

For atomic circular motion, combine

$$I = \frac{ev}{2\pi r}$$

and

$$B = \frac{\mu_0 I}{2r}$$

to get the direct result

$$B = \frac{\mu_0 e v}{4\pi r^2}$$

This works because the revolving charge behaves like a tiny current loop and the nucleus is at the centre of that loop.

Substitute

$$v = 6.76 \times 10^{6}\ m s^{-1}, \quad r = 0.52 \times 10^{-10}\ m$$

and constants $\mu_0$ and $e$ to obtain

$$B \approx 40\ T$$

Therefore, the required numerical value is $40$.