JEE Physics 2025 Question with Solution

Given:

• Magnetic field inside the solenoid is $B = 2.9 \times 10^{-4} \, \text{T}$
• Number of turns is $N = 200$
• Current is $I = 0.29 \, \text{A}$
• Permeability of free space is $\mu_0 = 4\pi \times 10^{-7} \, \text{T m/A}$

Find:

• Length of the solenoid, $l$

For a long solenoid, the magnetic field is

$$B = \mu_0 \frac{N}{l} I$$

Solving for $l$,

$$l = \frac{\mu_0 N I}{B}$$

Substituting the given values,

$$l = \frac{(4\pi \times 10^{-7})(200)(0.29)}{2.9 \times 10^{-4}} \, \text{m}$$ $$l = 8 \, \text{m}$$

Therefore, the length of the solenoid is $8 \, \text{m}$.

Given:

• A long solenoid has $N = 200$ turns
• It carries current $I = 0.29 \, \text{A}$
• Magnetic field inside it is $B = 2.9 \times 10^{-4} \, \text{T}$

Find:

• The solenoid length $l$

Using the relation

$$B = \mu_0 n I$$

where

$$n = \frac{N}{l}$$

So,

$$B = \mu_0 \frac{N}{l} I$$

Rearranging,

$$l = \frac{\mu_0 N I}{B}$$

Substitute the values from the given data,

$$l = \frac{(4\pi \times 10^{-7})(200)(0.29)}{2.9 \times 10^{-4}} \, \text{m}$$

This gives

$$l = 8 \, \text{m}$$

Hence, the required numerical value is 8.