A flat square coil of wire with 15 turns and an area of $0.40 \mathrm {~m} ^ { 2 }$ is placed with the plane of its area parallel to a magnetic field of $0.75 \mathrm {~T}$. The coil is flipped so its plane is perpendicular to the magnetic field in a time of $0.050 \mathrm {~s}$. What is the magnitude of the average induced emf in the coil?
A) $90 \mathrm {~V}$
B) $6.0 \mathrm {~V}$
C) $45 \mathrm {~V}$
D) $36 \mathrm {~V}$

Question

Quizplus · Accepted Answer

The average induced emf is given by:$$\varepsilon = -N\frac{\Delta \Phi}{\Delta t}$$where:* $N$ is the number of turns in the coil* $\Delta \Phi$ is the change in magnetic flux through the coil* $\Delta t$ is the time interval over which the change in flux occursIn this case, the initial magnetic flux through the coil is zero, since the plane of the coil is parallel to the magnetic field. When the coil is flipped, the magnetic flux through the coil changes by:$$\Delta \Phi = BA = (0.75 \mathrm {~T})(0.40 \mathrm {~m} ^ { 2 }) = 0.30 \mathrm {~Wb}$$The time interval over which the change in flux occurs is $0.050 \mathrm {~s}$. Therefore, the average induced emf is:$$\varepsilon = -N\frac{\Delta \Phi}{\Delta t} = -(15)(-0.30 \mathrm {~Wb}) / (0.050 \mathrm {~s}) = 90 \mathrm {~V}$$

A Flat Square Coil of Wire with 15 Turns and an Area