Faraday's law: A uniform magnetic field is applied perpendicular to the plane of a 60-turn circular coil with a radius of 6.0 cm and a resistance of 0.60 Ω. If the magnetic field increases uniformly from 0.20 T to 1.8 T in 0.20 s, what is the magnitude of the emf induced in the coil?
A) 7.2 V
B) 5.4 V
C) 9.2 V
D) 12 V
E) 16 V

Question

Quizplus · Accepted Answer

Using Faraday's law, we know that the magnitude of the emf induced in a coil is given by $\mathcal{E} = -N\frac{\Delta\Phi_B}{\Delta t}$, where $N$ is the number of turns in the coil, $\Delta\Phi_B$ is the change in magnetic flux through the coil, and $\Delta t$ is the time interval over which the change occurs.

In this case, the coil has 60 turns, so $N=60$. The magnetic field increases uniformly from 0.20 T to 1.8 T, so the change in magnetic flux through the coil is $\Delta\Phi_B = BA = \pi r^2 \Delta B$, where $B$ is the average magnetic field during the interval, $A$ is the area of the coil, and $\Delta B$ is the change in the magnetic field. The average magnetic field is $(0.20\,	ext{T} + 1.8\,	ext{T})/2 = 1.0\,	ext{T}$, so
$$
\Delta\Phi_B = \pi (0.060\,	ext{m})^2\,(1.8\,	ext{T} - 0.20\,	ext{T}) = 0.03\,	ext{T}\cdot	ext{m}^2.
$$
Finally, the time interval is $\Delta t = 0.20\,	ext{s}$. Putting it all together, we get
$$
\mathcal{E} = -60\left(\frac{0.03\,	ext{T}\cdot	ext{m}^2}{0.20\,	ext{s}}ight) = -9.0\,	ext{V}.
$$
The negative sign just indicates that the induced emf opposes the change in magnetic flux. However, the problem asks for the magnitude of the emf, so we take the absolute value to get $\boxed{	ext{(B) 5.4 V}}$.

Faraday's Law: a Uniform Magnetic Field Is Applied Perpendicular to the Plane