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Question

Physics Question on Electromagnetic induction

A circular ring of diameter 20cm20\,cm has a resistance of 0.010.01 Ω\Omega. The ring is turned from a position perpendicular to a uniform magnetic field of 2T2T to a position parallel to the field. The amount of charge flowing through the ring in this process is

A

πC\pi C

B

2πC2\pi C

C

8πC8\pi C

D

2C

Answer

2πC2\pi C

Explanation

Solution

When the ring is in perpendicular position, magnetic flux linked with it is maximum. i.e., ϕ1\phi_1 =BA = B×πr2=2×π×102=2π×102B \times \pi r^2 = 2 \times \pi \times 10^{-2} = 2\pi \times 10^{-2} Wb (r = 10 cm = 0.1 m) When ring is in parallel position, magnetic flux linked with it is zero. i.e., ϕ2\phi_2 = 0 \therefore Change in magnetic flux, Δϕ=ϕ1ϕ2=2π×102\Delta \phi = \phi_1 - \phi_2 = 2\pi \times 10^{-2} Wb Given, R = 0.01 Ω=102Ω\Omega = 10^{-2} \Omega Using, charge, q=ΔϕRq = \frac{\Delta \phi}{R} , we have, q=2π×102102=2πq = \frac{2 \pi \times 10^{-2}}{10^{-2}} = 2\pi C