NEET 2026 — Physics solution

NEET2026Physics

A 100-turn closely wound circular coil of radius 5 cm has a magnetic field of $3.14 \times 1 0^{- 3}$ T at its centre. The current flowing through the coil, and the magnitude of the magnetic moment of this coil are, respectively :

(Take $μ_{0} = 4 π \times 1 0^{- 7}$ T m/A)

A
2.5 A, 20 A m $^{2}$
B
2 A, 4 A m $^{2}$
C
2.5 A, 2 A m $^{2}$
D
2 A, 10 A m $^{2}$

Solution

For a circular coil, $B = \frac{μ _{0} N I}{2 R}$ , so

$I = \frac{2 B R}{μ _{0} N} = \frac{2 \times 3.14 \times 1 0 ^{- 3} \times 5 \times 1 0 ^{- 2}}{4 π \times 1 0 ^{- 7} \times 100} = 2.5$ A.

Magnetic moment $M = N I A = 100 \times 2.5 \times π (5 \times 1 0^{- 2})^{2} \approx 1.96 \approx 2$ A m $^{2}$ .

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