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The loop is in a magnetic field 0.20 T whose direction is perpendicular to the plane of the loop. At t = 0, the loop has area A = 0.285 square m.Suppose the radius of the elastic loop increases at a constant rate, dr/dt = 3.70 cm/s .

A) Determine the emf induced in the loop at t = 0.
B)Determine the emf induced in the loop at t = 1.00 s .

User Erikvm
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1 Answer

5 votes

Answer:

Part a)


EMF = 14 * 10^(-3) V

Part b)


EMF = 15.67 * 10^(-3) V

Step-by-step explanation:

As we know that magnetic flux through the loop is given as


\phi = B.A

now we have


\phi = B\pi r^2

now rate of change in flux is given as


(d\phi)/(dt) = B(2\pi r)(dr)/(dt)

now we know that


A = \pi r^2


0.285 = \pi r^2


r = 0.30 m

Now plug in all data


EMF = (0.20)* 2\pi* (0.30) * (0.037)


EMF = 14 * 10^(-3) V

Part b)

Now the radius of the loop after t = 1 s


r_1 = r_0 + (dr)/(dt)


r_1 = 0.30 + 0.037


r_1 = 0.337 m

Now plug in data in above equation


EMF = (0.20)* 2\pi* (0.337) * (0.037)


EMF = 15.67 * 10^(-3) V

User Hsmiths
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