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 perme - Physics MCQ NEET (UG) 2024 2026 | ExamTest.live

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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 units):

Solution & Step-by-step Explanation

The magnetic field at the centre of a circular coil with N turns is: B = \frac{μ _{0} N I}{2 R} Given: N = 100 I = 7 A R = 10 cm = 0.1 m μ_{0} = 4 π \times 1 0^{- 7} \approx 4 \times \frac{22}{7} \times 1 0^{- 7} $ B = \frac{4 \times \frac{22}{7} \times 1 0 ^{- 7} \times 100 \times 7}{2 \times 0.1} $ B = \frac{4 \times 22 \times 1 0 ^{- 5}}{0.2} = \frac{88 \times 1 0 ^{- 5}}{0.2} = 440 \times 1 0^{- 5} T B = 4.4 \times 1 0^{- 3} T = 4.4 mT

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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 units):

44 mT

4.4 T

4.4 mT

44 T

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