Question

A current of A is maintained in a circular loop of radius 14 cm. The value of dipole moment associated with the loop is

Answer 1

0.0616A A m^2

Answer 2

The magnetic dipole moment ($\mu$) of a current loop is given by the formula $\mu = NIA$, where $N$ is the number of turns, $I$ is the current, and $A$ is the area of the loop.

1. Identify given values:

   • The current is given as $I = A$ Amperes.
   • The loop is circular, so the number of turns $N=1$.
   • The radius of the loop is $R = 14$ cm.

2. Convert units: Convert the radius from centimeters to meters: $R = 14 \text{ cm} = 14 \times 10^{-2} \text{ m} = 0.14 \text{ m}$.

3. Calculate the area of the loop: The area of a circular loop is $A_{loop} = \pi R^2$. $A_{loop} = \pi (0.14 \text{ m})^2 = \pi (0.0196) \text{ m}^2$.

4. Calculate the magnetic dipole moment: Substitute the values into the formula $\mu = NIA$. $\mu = 1 \times A \times (0.0196\pi) \text{ A m}^2$. $\mu = 0.0196\pi A \text{ A m}^2$.

5. Approximate using $\pi \approx \frac{22}{7}$: $\mu = 0.0196 \times \frac{22}{7} \times A \text{ A m}^2$. $\mu = (0.0028 \times 22) \times A \text{ A m}^2$. $\mu = 0.0616A \text{ A m}^2$.

The value of the dipole moment associated with the loop is $0.0616A \text{ A m}^2$.