A rectangular wire loop (length = 60 cm, width = 40 cm) lies completely within a perpendicular and uniform magnetic field of magnitude of 0.5 T. If the length of the loop starts increasing at a rate of 20 mm/s at time t = 0, while the width is decreasing at the same rate, what is the magnitude of the induced emf at time t = 4.0 s?
A) 6.8 mV
B) 5.2 mV
C) 3.6 mV
D) 8.4 mV
E) 10 mV

Question

Quizplus · Accepted Answer

The area enclosed by the loop is A = lw = (0.6 m)(0.4 m) = 0.24 m^2.

The rate of change of the length and the width are both given as dL/dt = dW/dt = 20 mm/s = 0.02 m/s.

The induced emf in the loop is given by:

emf = -d/dt (BA cosθ)

where B is the magnitude of the magnetic field, A is the area enclosed by the loop, and θ is the angle between the magnetic field and the normal to the loop.

In this case, the angle θ is 90 degrees since the loop is perpendicular to the magnetic field. Therefore,

emf = -d/dt (BA cos90) = -d/dt (BA)

Since the loop is rectangular, the area vector is perpendicular to the magnetic field and has a magnitude equal to the area of the loop. Therefore,

emf = -d/dt (B A^2)

Substituting in the given values, we get

emf = -(0.5 T)(0.24 m^2)(2 × 10^-4 m/s) = -0.024 V

The negative sign indicates that the induced emf creates a current in the loop that opposes the change in the magnetic flux.

At t = 4.0 s, the total change in the area is

ΔA = (dL/dt)(t) - (dW/dt)(t) = (0.02 m/s)(4.0 s) - (0.02 m/s)(4.0 s) = 0

Therefore, the induced emf at t = 4.0 s is still -0.024 V or 24 mV. However, we are asked for the magnitude of the induced emf, which is a positive quantity. Therefore, the answer is

Answer: C (3.6 mV)

A Rectangular Wire Loop (Length = 60 Cm, Width =