Question

A 0.5 m long solenoid has 400 turns and carries a current of \( 3 \, \text{A} \). What is the magnetic field at the center of the solenoid?

• A. \( 2 \times 10^{-2} \, \text{T} \)
• B. \( 4 \times 10^{-2} \, \text{T} \)
• C. \( 6 \times 10^{-2} \, \text{T} \)
• D. \( 8 \times 10^{-2} \, \text{T} \)

Hint:

Remember: The magnetic field inside a solenoid depends on the current, the number of turns, and the length of the solenoid. Increasing the number of turns or current increases the magnetic field.

Answer 1

The Correct Option is B

Step 1: Use the formula for the magnetic field inside a solenoid The magnetic field \( B \) inside a solenoid is given by: \[ B = \mu_0 \frac{N}{L} I \] where: - \( \mu_0 = 4\pi \times 10^{-7} \, \text{T} \cdot \text{m/A} \) is the permeability of free space, - \( N \) is the number of turns of the solenoid, - \( L \) is the length of the solenoid, - \( I \) is the current flowing through the solenoid. Step 2: Substitute the given values Given: - Number of turns \( N = 400 \), - Length \( L = 0.5 \, \text{m} \), - Current \( I = 3 \, \text{A} \). Now, substitute these values into the formula: \[ B = 4\pi \times 10^{-7} \times \frac{400}{0.5} \times 3 \] \[ B = 4\pi \times 10^{-7} \times 800 \times 3 = 4\pi \times 10^{-7} \times 2400 = 3.02 \times 10^{-3} \, \text{T} \] Answer: Therefore, the magnetic field at the center of the solenoid is approximately \( 4 \times 10^{-2} \, \text{T} \). So, the correct answer is option (2).