A long solenoid has 2 × 10^4 turns per meter. At the centre, a coil of 100 turns and radius 0.01 m is placed. The current in the solenoid reduces to 0 - Physics MCQ NEET (UG) 2017 2026 | ExamTest.live

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A long solenoid has 2 \times 1 0^{4} turns per meter. At the centre, a coil of 100 turns and radius 0.01 m is placed. The current in the solenoid reduces to 0 A from 4 A in 0.05 s . If the resistance of the coil is 10 π^{2} Ω, the total charge flowing through the coil is:

Solution & Step-by-step Explanation

Total charge q = \frac{ΔΦ}{R} = \frac{N Δ B A}{R} . Δ B = μ_{0} n Δ I = (4 π \times 1 0^{- 7}) \times (2 \times 1 0^{4}) \times 4 . Δ B = 32 π \times 1 0^{- 3} T . A = π r^{2} = π (0.01)^{2} = π \times 1 0^{- 4} m^{2} . q = \frac{100 \times ( 32 π \times 1 0 ^{- 3} ) \times ( π \times 1 0 ^{- 4} )}{10 π ^{2}} q = \frac{100 \times 32 π ^{2} \times 1 0 ^{- 7}}{10 π ^{2}} = 320 \times 1 0^{- 7} = 32 \times 1 0^{- 6} C = 32 μ C

Try it yourself before checking the explanation above.

A long solenoid has 2 \times 1 0^{4} turns per meter. At the centre, a coil of 100 turns and radius 0.01 m is placed. The current in the solenoid reduces to 0 A from 4 A in 0.05 s . If the resistance of the coil is 10 π^{2} Ω, the total charge flowing through the coil is:

16 μ C

32 μ C

16 π μ C

32 π μ C

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