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. \( 3\times 10^{-2} \, \text{T} \)
• 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

We are given the following data:

• Length of the solenoid \( L = 0.5 \, \text{m} \)
• Number of turns \( N = 400 \)
• Current \( I = 3 \, \text{A} \)

Step 1: Recall the formula for the magnetic field inside a solenoid

The magnetic field \( B \) inside a solenoid is given by the formula: \[ B = \mu_0 \cdot \frac{N}{L} \cdot I \]

Step 2: Substitute the values into the formula

\[ B = (4\pi \times 10^{-7}) \cdot \frac{400}{0.5} \cdot 3 \]

Step 3: Simplify the calculation

\[ B = (4\pi \times 10^{-7}) \cdot 800 \cdot 3 \] \[ B = 3.0 \times 10^{-2} \, \text{T} \]

Conclusion:

The magnetic field at the center of the solenoid is \( 3.0 \times 10^{-2} \, \text{T} \).