A circular loop is being pulled out with a constant
speed out of a region of uniform magnetic field as
shown. The induced emf in the loop (R = Radius
of loop)
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Answer:
please mention your diagram please
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Hence the electrical power in the loop is P = B^2 a^2 v^2 / R
Explanation:
We have the area enclosed by the magnetic field at any instant as
A = πa^2 − [πa^2 ( 2θ/ 2π ) − x / 2 .2 a sinθ]
A = πa 2 ^−a 2 θ + axsinθ
A = πa^2 − a^ 2 cos −1 ( x/a )+ax a x 2 −a 2
Thus we get the emf as ϵ = dt / dϕ = B dA / dt
We calculate dA/dt
= 0 + a^2 √1 - x^ 2 / a^2 1/a dx / dt + [ √a^2 - x^2 - x / 2 2x /√x^2 - a^2]dx /dt
θ = π / 3 and x = π √ 3 a / 2
Solving this we get dA / dt = a v
P = ϵ^2 / R = B^2 / R (dA / dt)^2
P = B^2 a^2 v^2 / R
Hence the electrical power in the loop is P = B^2 a^2 v^2 / R
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