Question

A long solenoid of length \(75\) cm carries a current of \(3.5\) A. If the number of turns of the solenoid is \(600\), the magnitude of the magnetic field inside the solenoid is

• A. \(7 \times 10^{-3} \, T\)
• B. \(2.48 \times 10^{-3} \, T\)
• C. \(3 \times 10^{-3} \, T\)
• D. \(3.52 \times 10^{-3} \, T\)

Hint:

The magnetic field inside a solenoid depends on the number of turns per unit length and the current passing through it. Always use the formula: \[ B = \mu_0 n I, \quad \text{where } n = \frac{N}{L} \] Ensure that all units are in SI before calculation.

Answer 1

The Correct Option is D

- The magnetic field inside a long solenoid is given by: \[ B = \mu_0 n I \]
- Given:
- \( \mu_0 = 4\pi \times 10^{-7} \, T\,m/A \)
- \( N = 600 \)
- \( L = 0.75 \, m \)
- \( I = 3.5 \, A \)
- First, find number of turns per unit length: \[ n = \frac{N}{L} = \frac{600}{0.75} = 800 \, m^{-1} \]
- Now substitute in the formula: \[ B = (4\pi \times 10^{-7}) \times 800 \times 3.5 \]
- Simplify: \[ B = (4\pi \times 10^{-7}) \times 2800 \]
- Using \( \pi \approx 3.14 \): \[ B = (1.256 \times 10^{-6}) \times 2800 \]
- Final value: \[ B = 3.52 \times 10^{-3} \, T \]
- Hence, the magnetic field is: 3.52 × 10⁻³ T