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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 .

Respuesta :

Answer:

Part a)

[tex]EMF = 14 \times 10^{-3} V[/tex]

Part b)

[tex]EMF = 15.67 \times 10^{-3} V[/tex]

Explanation:

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

[tex]\phi = B.A[/tex]

now we have

[tex]\phi = B\pi r^2[/tex]

now rate of change in flux is given as

[tex]\frac{d\phi}{dt} = B(2\pi r)\frac{dr}{dt}[/tex]

now we know that

[tex]A = \pi r^2[/tex]

[tex]0.285 = \pi r^2[/tex]

[tex]r = 0.30 m[/tex]

Now plug in all data

[tex]EMF = (0.20)\times 2\pi\times (0.30) \times (0.037)[/tex]

[tex]EMF = 14 \times 10^{-3} V[/tex]

Part b)

Now the radius of the loop after t = 1 s

[tex]r_1 = r_0 + \frac{dr}{dt}[/tex]

[tex]r_1 = 0.30 + 0.037[/tex]

[tex]r_1 = 0.337 m[/tex]

Now plug in data in above equation

[tex]EMF = (0.20)\times 2\pi\times (0.337) \times (0.037)[/tex]

[tex]EMF = 15.67 \times 10^{-3} V[/tex]

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