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

Step 1: Using the Formula for Magnetic Field in a Solenoid The magnetic field inside a long solenoid is given by: \[ B = \mu_0 n I \] where:
- \( \mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A \) (permeability of free space),
- \( n = \frac{N}{L} \) is the number of turns per unit length,
- \( I = 3.5 \) A (current through the solenoid),
- \( N = 600 \) (total number of turns),
- \( L = 75 \) cm \(= 0.75\) m (length of the solenoid). Step 2: Substituting Values First, we calculate \( n \): \[ n = \frac{N}{L} = \frac{600}{0.75} = 800 \, \text{turns/m} \] Now, we compute \( B \): \[ B = (4\pi \times 10^{-7}) \times (800) \times (3.5) \] \[ B = (4\pi \times 10^{-7}) \times 2800 \] \[ B = (4 \times 3.1416 \times 10^{-7}) \times 2800 \] \[ B = (1.256 \times 10^{-6}) \times 2800 \] \[ B = 3.52 \times 10^{-3} \, T \] Thus, the correct answer is \( \mathbf{(4)} \ 3.52 \times 10^{-3} \, T \).