Mastering Physics Solutions: Exercise 20.7

Exercise 20.7

Part A = 1.7 V Click to use the calculator/solver for this part of the problem
Part B = 0.44 V Click to use the calculator/solver for this part of the problem

Solutions Below:

A square loop of wire with sides of length 40 cm is in a uniform magnetic field perpendicular to its area.

Part A

If the field’s strength is initially 120 mT and it decays to zero in 0.011 s, what is the magnitude of the average emf induced in the loop?
Express your answer using two significant figures.

The induced emf is just the change in magnetic flux per unit time. Start by finding the magnetic flux through the loop:

ΔΦ = A * ΔB

Where A is the area of the loop (0.40 m * 0.40 m = 0.16 m2)

So:

ΔΦ = 0.16 * 0.12
ΔΦ = 0.0192

The formula for emf (ε) is:

ε = ΔΦ / Δt
ε = 0.0192 / 0.011

1.7 V

Part B

What would be the average emf if the sides of the loop were only 20 cm?
Express your answer using two significant figures.

Just use the same formula as in Part A:

ΔΦ = A * ΔB
ΔΦ = (0.2 * 0.2) * 0.12
ΔΦ = 0.0048

And:

ε = ΔΦ / Δt
ε = 0.0048 / 0.011

0.44 V

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