# Induced EMF and Current in a Shrinking Loop?

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A circular loop of flexible iron wire has an initial circumference of 165 cm , but its circumference is decreasing at a constant rate of 14.0 cm/s due to a tangential pull on the wire. The loop is in a constant uniform magnetic field of magnitude 0.900 T , which is oriented perpendicular to the plane of the loop. Assume that you are facing the loop and that the magnetic field points into the loop.

Find the magnitude of the emf induced in the loop after exactly time 2.00 s has passed since the circumference of the loop started to decrease.

• Circumference C= 2πr

Rate of change of circumference is

dC/dt = 2π dr /dt = – 0.14m/s ( GIVEN)

Minus sign to denote the decrease in value.

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A = πr^2

The rate of change of area

dA/dt = 2πr dr /dt = – 0.14*r (Replacing 2π dr /dt by -0.14)

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Initial circumference = 1.65 m

In 2 second the C decreases by 2*0.14 = 0.28 m

The radius after 2 s is [1.65 – 0.28] / (2π) = 0.22 m

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e = – d/dt {BA cos θ]

e = – 0.9* dA/dt since cos θ = 1 and B = 0.9

e = – 0.9 *dA/dt

e = 0.9 *[0.14*r] (Substituting for dA/dt as – 0.14*r)

e = 0.9 *0.14*0.22 (Substituting the value of r)

e = 0.028 V

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• A circular loop of flexible iron wire has an initial circumference of 165 cm , but its circumference is decreasing at a constant rate of 14.0 cm/s due to a tangential pull on the wire. The loop is in a constant uniform magnetic field of magnitude 0.900 T , which is oriented perpendicular to the plane of the loop. Assume that you are facing the loop and that the magnetic field points into the loop.

Find the magnitude of the emf induced in the loop after exactly time 2.00 s has passed since the circumference of the loop started to decrease.

EMF= magnetic field * area

The derivative of this equation with respect to time is the change in EMF with respect to time

Eq.#1. = dEMF / dt = B * dA / dt

As the area changes, the EMF changes.

We need an equation that expresses the area in terms of the circumference.

C = 2 * π * r and Area = π * r^2

r = C ÷ (2 * π)

Area = π * (C ÷ 2 * π)^2

Eq. #2 = Area = C^2 ÷ (4 * π)

The derivative of this function with respect to time is the change in area with respect to time

dA/dt = d(C^2 ÷ 4 π)/dt = (2 C/4 π )* dC / dt

dA/dt =(C/2 π )* dC / dt

Now substitute this expression for dA / dt into Eq.#1

Eq.#1. = dEMF / dt = B * dA / dt

Eq.#1. = dEMF / dt = B * (C/2 π )* dC / dt

C = 1.65 m

B = 0.9 T

dC / dt = 0.14 m/s

At 2 seconds, C = 1.65 – (0.14 * 2) = 1.37 m

dEMF / dt = B * (C/2 π )* dC / dt

dEMF / dt = 0.9 * (1.37/2 π) * 0.14 = 0.02747 