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Step2247 [10]
3 years ago
12

A 5.2 cm diameter circular loop of wire is in a 1.35 T magnetic field. The loop is removed from the field in 0.29 sec. Assume th

at the loop is perpendicular to the magnetic field. What is the average induced emf?
Physics
1 answer:
Katyanochek1 [597]3 years ago
3 0

Answer:

9.88 milivolt

Explanation:

Given: diameter d = 5.2 cm

magnetic field B_1 = 1.35 T, final magnetic field B_2 =0 T

t = 0.29 sec.

we know emf =  - dΦ/dt

and flux Φ = BA

A= area

therefore emf ε = -A(B_2-B_1)/Δt

=-\pi(d/2)^2\frac{B_2-B_1}{\Delta t} \\=-\pi(0.052/2)^2\frac{0-1.35}{0.29} \\=98.8\times10^4\\=9.88 mV

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The force exerted on the board by the karate master given the data is -4500 N

<h3>Data obtained from the question </h3>
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