A long solenoid of 50 cm length having 100 turns carries a current of 2.5 A . The magnetic field at the centre of the solenoid is:( μ_0 = 4π × 10^-7 T - Physics MCQ NEET 2020 E1 2026 | ExamTest.live

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A long solenoid of 50 cm length having 100 turns carries a current of 2.5 A . The magnetic field at the centre of the solenoid is:(μ_{0} = 4 π \times 1 0^{- 7} T m A^{- 1})

Solution & Step-by-step Explanation

Magnetic field inside a solenoid B = μ_{0} n I = μ_{0} (N / L) I .Given N = 100, L = 0.5 m, I = 2.5 A . B = 4 π \times 1 0^{- 7} \times \frac{100}{0.5} \times 2.5 B = 4 π \times 1 0^{- 7} \times 200 \times 2.5 = 4 π \times 1 0^{- 7} \times 500 = 2000 π \times 1 0^{- 7} B = 2 π \times 1 0^{- 4} \approx 6.28 \times 1 0^{- 4} T

Try it yourself before checking the explanation above.

A long solenoid of 50 cm length having 100 turns carries a current of 2.5 A . The magnetic field at the centre of the solenoid is:(μ_{0} = 4 π \times 1 0^{- 7} T m A^{- 1})

6.28 \times 1 0^{- 4} T

3.14 \times 1 0^{- 4} T

6.28 \times 1 0^{- 5} T

3.14 \times 1 0^{- 5} T

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