Question

A coil of 200 turns and area of cross-section \(5 \times 10^{-3} \, m^2\) carries a current of 3 A. - The magnetic moment of the coil is

• A. \(12 \, Am^2\)
• B. \(3 \, Am^2\)
• C. \(6 \, Am^2\)
• D. \(9 \, Am^2\)

Hint:

The magnetic moment of a coil is given by: \[ M = N I A \] Always check the number of turns, current, and cross-sectional area when solving magnetic moment problems.

Answer 1

The Correct Option is B

Step 1: Using the Formula for Magnetic Moment of a Coil The magnetic moment \( M \) of a current-carrying coil is given by: \[ M = N I A \] where: 
- \( N = 200 \) (number of turns), 
- \( I = 3 \) A (current in the coil), 
- \( A = 5 \times 10^{-3} \, m^2 \) (area of cross-section). 

Step 2: Substituting Values \[ M = (200) \times (3) \times (5 \times 10^{-3}) \] \[ M = 200 \times 15 \times 10^{-3} \] \[ M = 3 \, Am^2 \] Thus, the correct answer is \( \mathbf{(2)} \ 3 \, Am^2 \).