A ten-loop coil having an area of 0.23 m2 and a very large resistance is in a 0.047-T uniform magnetic field oriented so that the maximum flux goes through the coil. The coil is then rotated so that the flux through it goes to zero in 0.34 s. What is the magnitude of the average emf induced in the coil during the 0.34 s? A ten-loop coil having an area of 0.23 m2 and a very large resistance is in a 0.047-T uniform magnetic field oriented so that the maximum flux goes through the coil. The coil is then rotated so that the flux through it goes to zero in 0.34 s. What is the magnitude of the average emf induced in the coil during the 0.34 s? 0.00 V 0.0032 V 0.032 V 0.32 V 1.0 V

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Luka

Physics

11 September, 05:58

A ten-loop coil having an area of 0.23 m2 and a very large resistance is in a 0.047-T uniform magnetic field oriented so that the maximum flux goes through the coil. The coil is then rotated so that the flux through it goes to zero in 0.34 s. What is the magnitude of the average emf induced in the coil during the 0.34 s? A ten-loop coil having an area of 0.23 m2 and a very large resistance is in a 0.047-T uniform magnetic field oriented so that the maximum flux goes through the coil. The coil is then rotated so that the flux through it goes to zero in 0.34 s. What is the magnitude of the average emf induced in the coil during the 0.34 s? 0.00 V 0.0032 V 0.032 V 0.32 V 1.0 V

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Hopkins · Answer 1

The magnitude of the average emf induced in the coil is approximately o. 32 V.

Explanation:

As, no. of loops = N = 10

Area = 0.23 m^2

Magnetic field = B = 0.047 T

Δt = 0.34 s

These are the given things now we have to find the EMF. So the formula of emf is:

emf = e = N*ΔФ/Δt

Therefore, ΔФ = B. ΔA = BΔAcos (θ)

Since, flux is maximum so θ = 0 and cos (0) = 1

Then,

ΔФ = BA

e = N * (BΔA) / Δt

e = 10 * (0.047*0.23) / 0.34

e = 0.317 V.

So, we can write it by rounding off the digit

e ≅ 0.32 V.