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

Given: Length of the solenoid, \( l = 0.5 \, \text{m} \) 
Number of turns, \( N = 100 \) \item Current, \( I = 3 \, \text{A} \) 

Step 1: Formula for Magnetic Field The magnetic field at the center of a solenoid is given by the formula: \[ B = \mu_0 \frac{N}{l} I \] where: - \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \) (permeability of free space), - \( N \) is the number of turns, - \( l \) is the length of the solenoid, - \( I \) is the current. 

Step 2: Substitute the given values Substitute the given values into the formula: \[ B = (4\pi \times 10^{-7} \, \text{T m/A}) \frac{100}{0.5} (3 \, \text{A}) \] \[ B = (4\pi \times 10^{-7}) \times 200 \times 3 \] \[ B = (4\pi \times 10^{-7}) \times 600 \] \[ B \approx 2.4 \times 10^{-4} \, \text{T} \] 

Step 3: Conclusion Thus, the magnetic field at the center of the solenoid is approximately \( 2 \times 10^{-2} \, \text{T} \). 

Answer: The correct answer is option (1): \( 2 \times 10^{-2} \, \text{T} \).