Question

A current of \( \frac{10}{\pi} \) A is maintained in a circular loop of radius 14 cm. The value of dipole moment associated with the loop is:

• A. 0.019 Am\(^2\)
• B. 0.14 Am\(^2\)
• C. 0.196 Am\(^2\)
• D. 0.615 Am\(^2\)

Hint:

To find the magnetic dipole moment of a current loop, use the formula \( \mu = I \cdot A \), where \( A \) is the area of the loop. For a circular loop, \( A = \pi r^2 \).

Answer 1

The Correct Option is C

The magnetic dipole moment \( \mu \) of a current-carrying loop is \( \mu = I \cdot A \), where \( I \) is the current and \( A \) is the loop's area. For a circular loop, \( A = \pi r^2 \), with \( r \) being the radius. Given \( I = \frac{10}{\pi} \, \text{A} \) and \( r = 14 \, \text{cm} = 0.14 \, \text{m} \), the area is \( A = \pi (0.14)^2 = \pi \times 0.0196 = 0.0616 \, \text{m}^2 \). The magnetic dipole moment is then \( \mu = \frac{10}{\pi} \times 0.0616 = 0.196 \, \text{Am}^2 \). Therefore, the dipole moment is \( 0.196 \, \text{Am}^2 \), corresponding to option (C).