Mastering Physics Solutions: Exercise 20.6

Exercise 20.6

Part A = 0.33 m2 Click to use the calculator/solver for this part of the problem
Part B = 42 V Click to use the calculator/solver for this part of the problem
Part C = 83 V Click to use the calculator/solver for this part of the problem

Solutions Below:

A uniform magnetic field is at right angles to the plane of a wire loop. The field decreases by 0.25 T in 1.0*10^−3 s and the magnitude of the average emf induced in the loop is 83 V.

Part A

What is the area of 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 = A * 0.25

The formula for emf (ε) is:

ε = ΔΦ / Δt

Substitute in for ΔΦ and the voltage and time that the problem gave us:

83 = (A * 0.25) / 0.001
83 = 250 * A
A = 0.332 m2

Round to 2 significant figures:

0.33 m2

Part B

What would be the value of the average induced emf if the field change was the same but took twice as long to decrease?
Express your answer using two significant figures.

Just use the same formula as in Part A:

ε = ΔΦ / Δt
ε = (A * 0.25) / 0.002
ε = (0.332 * 0.25) / 0.002
ε = 41.5

Note that mastering physics is picky (as usual) about this problem. If you used the 0.33 m2 for area that you found in Part A, you’ll get the wrong answer. You have to plug in the unrounded 0.332 m2, get 41.5 V, and then round up to 42 V.

42 V

Part C

What would be the value of the average induced emf if the field decrease was twice as much and it also took twice as long to change?
Express your answer using two significant figures.

Just use the same formula as in Part A:

ε = ΔΦ / Δt
ε = (0.332 * 0.50) / 0.002
ε = 83

83 V

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