Question

A 0.5 m long solenoid has 100 turns and carries a current of 3A. 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 is given by \( B = \mu_0 \frac{N}{l} I \). Ensure that you use the correct units for all quantities.

Answer 1

The Correct Option is A

Magnetic Field at the Center of a Solenoid

Given data:

• Solenoid length, \( L = 0.5 \, \text{m} \)
• Number of turns, \( N = 100 \)
• Current, \( I = 3 \, \text{A} \)
• Permeability of free space, \( \mu_0 = 4\pi \times 10^{-7} \, \text{T} \cdot \text{m/A} \)

Step 1: Formula for magnetic field inside a solenoid:

\[ B = \mu_0 \cdot \frac{N}{L} \cdot I \]

Step 2: Substitute values:

\[ B = (4\pi \times 10^{-7}) \cdot \frac{100}{0.5} \cdot 3 \]

Step 3: Simplify:

\[ B = 2.4 \times 10^{-4} \, \text{T} \]

Conclusion:

The magnetic field at the center of the solenoid is \( 2.4 \times 10^{-4} \, \text{T} \).