NEET 2024 — Physics solution

NEET2024Physics

A tightly wound 100 turns coil of radius 10 cm carries a current of 7 A. The magnitude of the magnetic field at the centre of the coil is (Take permeability of free space as $4 π \times 1 0^{- 7}$ SI unit):

A
4.4 mT
B
44 T
C
44 mT
D
4.4 T

Solution

Field at the centre of a circular coil: $B = \frac{μ _{0} N I}{2 R}$ .

$B = \frac{4 π \times 1 0 ^{- 7} \times 100 \times 7}{2 \times 0.1} = \frac{4 π \times 7 \times 1 0 ^{- 5}}{0.2} = 14 π \times 1 0^{- 4}$ T $\approx 4.4 \times 1 0^{- 3}$ T $= 4.4$ mT.

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