JEE Physics 2025 Question with Solution

Given: current in the solenoid is $I = 1.5 \, \text{A}$ and the time period of the electron is $T = 75 \times 10^{-9} \, \text{s}$.

Find: the number of turns per meter $n$ in the solenoid.

Inside a long solenoid, the magnetic field is

$$B = \mu_0 n I$$

For the electron in uniform circular motion, magnetic force provides the centripetal force:

$$e v B = \frac{m v^2}{r}$$

Using

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

we get

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

Substituting in the force relation,

$$B = \frac{m v}{e r} = \frac{m}{e r}\cdot \frac{2\pi r}{T} = \frac{2\pi m}{eT}$$

Now substitute $m = 9.11 \times 10^{-31} \, \text{kg}$, $e = 1.6 \times 10^{-19} \, \text{C}$ and $T = 75 \times 10^{-9} \, \text{s}$:

$$B = \frac{2\pi \times 9.11 \times 10^{-31}}{1.6 \times 10^{-19} \times 75 \times 10^{-9}} = 1.58 \times 10^{-3} \, \text{T}$$

Using

$$B = \mu_0 n I$$

with $\mu_0 = 4\pi \times 10^{-7} \, \text{T m/A}$ and $I = 1.5 \, \text{A}$,

$$1.58 \times 10^{-3} = (4\pi \times 10^{-7}) n \times 1.5$$

So,

$$n = \frac{1.58 \times 10^{-3}}{(4\pi \times 10^{-7}) \times 1.5} = 250$$

Therefore, the number of turns per meter in the solenoid is $250$.

Given: $I = 1.5 \, \text{A}$ and $T = 75 \times 10^{-9} \, \text{s}$.

Find: $n$.

For a charged particle moving perpendicular to a uniform magnetic field, the time period is directly

$$T = \frac{2\pi m}{eB}$$

Hence,

$$B = \frac{2\pi m}{eT}$$

For a long solenoid,

$$B = \mu_0 n I$$

Therefore,

$$n = \frac{2\pi m}{\mu_0 e T I}$$

Substituting the values,

$$n = \frac{2\pi \times 9.11 \times 10^{-31}}{(4\pi \times 10^{-7})(1.6 \times 10^{-19})(75 \times 10^{-9})(1.5)} \approx 250$$

This shortcut works because the orbital time period in a magnetic field is independent of the radius and speed. Therefore, the required number of turns per meter is $250$.