Question

A 100-turn closely wound circular coil of radius 5 cm has a magnetic field of 3.14 $\times$ 10⁻³ T at its centre. The current flowing through the coil, and the magnitude of the magnetic moment of this coil are, respectively: (Take $\mu_0 = 4\pi \times 10^{-7}$ T m/A)

• A. 2 A, 4 A m²
• B. 2.5 A, 20 A m²
• C. 2.5 A, 2 A m²
• D. 2 A, 10 A m²

Hint:

In competitive exams, $3.14$ is often used interchangeably with $\pi$. Canceling $\pi$ on both sides of your equations early on usually makes the calculation much simpler.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A current-carrying coil creates a magnetic field at its center and also acts as a magnetic dipole with a specific magnetic moment.

Step 2: Key Formula or Approach:
1. Magnetic field at centre ($B$) = $\frac{\mu_0 NI}{2r}$ 2. Magnetic moment ($M$) = $NIA$, where $A = \pi r^2$

Step 3: Detailed Explanation:
Given: $N=100$, $r=0.05$ m, $B=3.14 \times 10^{-3}$ T (which is $\pi \times 10^{-3}$ T). 1. Find Current ($I$): \[ 3.14 \times 10^{-3} = \frac{4\pi \times 10^{-7} \times 100 \times I}{2 \times 0.05} \] \[ \pi \times 10^{-3} = \frac{4\pi \times 10^{-5} \times I}{0.1} \] \[ 10^{-3} = 4 \times 10^{-4} \times I \] \[ I = \frac{10^{-3}}{4 \times 10^{-4}} = \frac{10}{4} = 2.5 \text{ A} \] 2. Find Magnetic Moment ($M$): \[ A = \pi r^2 = \pi \times (0.05)^2 = 3.14 \times 0.0025 \text{ m}^2 \] \[ M = 100 \times 2.5 \times (3.14 \times 0.0025) \] \[ M = 250 \times 0.00785 \approx 1.96 \approx 2 \text{ A m}^2 \]

Step 4: Final Answer:
The current is 2.5 A and the magnetic moment is 2 A m².