Question

A wire of length 314 cm carrying current of 14 A is bent to form a circle. The magnetic moment of the coil is _______ A–m². [Given \(\pi = 3.14\)]

Answer 1

Correct Answer: 11

To find the magnetic moment of a wire bent to form a circle, we first need the formula for the magnetic moment \( m \) of a coil: \( m = n \cdot I \cdot A \), where \( n \) is the number of turns, \( I \) is the current, and \( A \) is the area of the circle.
1. The wire length is given as 314 cm. When bent into a circle, the circumference \( C \) of the circle is the length of the wire, so \( C = 2\pi r \).
2. Solving for the radius \( r \) gives: \( r = \frac{C}{2\pi} = \frac{314}{2 \times 3.14} = 50 \) cm.
3. Convert radius into meters: \( 50 \) cm = \( 0.5 \) m. 
4. Calculate the area \( A \) of the circle: \( A = \pi r^2 = 3.14 \times (0.5)^2 = 3.14 \times 0.25 = 0.785 \) m².
5. Since it's a single loop, \( n = 1 \).
6. The current \( I \) is given as 14 A.
7. Substitute the values to find the magnetic moment: \( m = 1 \times 14 \times 0.785 = 10.99 \, \text{A-m}^2 \).
8. Given range is from 11 to 11, hence \( 10.99 \) is approximately equal to 11 A-m² which is within the accepted range.

Answer 2

\(R = \frac{I}{2\pi}\)
\(\frac{314}{2 \times 3.14}\)
\(=50 cm\)
The magnetic moment of the coil,
\(\mu = \pi R^2 i\)
\(14 \times 3.14 \times (0.5)^2\)
\(= 11\) A-m²
So, the answer is 11.