Q136 / 180 · NEET 2026 (Cancelled) · Physics
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)
(Take $\mu_0 = 4\pi \times 10^{-7}$ T m/A)
- A$2$ A, $10$ A m$^2$
- B$2.5$ A, $20$ A m$^2$
- C$2$ A, $4$ A m$^2$
- Correct answerD$2.5$ A, $2$ A m$^2$
Solution
The magnetic field at the centre of a circular coil is given by:
$B = \dfrac{\mu_0 N I}{2R}$
Substituting the given values ($B = 3.14 \times 10^{-3}$ T, $N = 100$, $R = 0.05$ m, $\mu_0 = 4\pi \times 10^{-7}$ T m/A):
$3.14 \times 10^{-3} = \dfrac{4\pi \times 10^{-7} \times 100 \times I}{2 \times 0.05}$
Taking $\pi = 3.14$:
$3.14 \times 10^{-3} = \dfrac{4 \times 3.14 \times 10^{-5} \times I}{0.1}$
$10^{-3} = 4 \times 10^{-4} \times I$
$I = \dfrac{10^{-3}}{4 \times 10^{-4}} = 2.5 \text{ A}$
The magnetic moment of the coil is given by:
$M = N I A = N I (\pi R^2)$
$M = 100 \times 2.5 \times 3.14 \times (0.05)^2$
$M = 250 \times 3.14 \times 0.0025$
$M = 1.9625 \text{ A m}^2 \approx 2 \text{ A m}^2$
Answer: $2.5$ A, $2$ A m$^2$
$B = \dfrac{\mu_0 N I}{2R}$
Substituting the given values ($B = 3.14 \times 10^{-3}$ T, $N = 100$, $R = 0.05$ m, $\mu_0 = 4\pi \times 10^{-7}$ T m/A):
$3.14 \times 10^{-3} = \dfrac{4\pi \times 10^{-7} \times 100 \times I}{2 \times 0.05}$
Taking $\pi = 3.14$:
$3.14 \times 10^{-3} = \dfrac{4 \times 3.14 \times 10^{-5} \times I}{0.1}$
$10^{-3} = 4 \times 10^{-4} \times I$
$I = \dfrac{10^{-3}}{4 \times 10^{-4}} = 2.5 \text{ A}$
The magnetic moment of the coil is given by:
$M = N I A = N I (\pi R^2)$
$M = 100 \times 2.5 \times 3.14 \times (0.05)^2$
$M = 250 \times 3.14 \times 0.0025$
$M = 1.9625 \text{ A m}^2 \approx 2 \text{ A m}^2$
Answer: $2.5$ A, $2$ A m$^2$
Hence, the correct answer is option: 'D'.