Question

Figure a shows two concentric circular regions in which uniform magnetic fields can change. Region 1, with radius $$r_1=1.0cm$$, has an outward magnetic field $$\overrightarrow {B_1}$$ that is increasing in magnitude. Region 2, with radius $$r_2=2.0cm$$, has an outward magnetic field $$\overrightarrow {B_2}$$ that may also be changing.
Imagine that a conducting ring of radius R is centered on the two regions and then the emf $$\mathscr{E}$$ around the ring is determined. Figure b gives emf $$\mathscr{E}$$ as a function of the square $$R^2$$ of the rings radius, to the outer edge of region 2. The vertical axis scale is set by $$\mathscr{E}_s=20.0nV$$. What are the rates (a) $$dB_1/dt$$ and (b) $$dB_2/dt$$? (c) Is the magnitude of $$\overrightarrow {B}_2$$increasing, decreasing, or remaining constant?

Answer 1

In absolute value, Faraday's law (for a single turn, with $$B$$ changing in time) gives

$$\dfrac{d
\Phi_{B}}{d t}=\dfrac{d(B A)}{d t}=A \dfrac{d B}{d t}=\pi R^{2} \dfrac{d B}{d
t}$$

for the magnitude of the induced emf. Dividing it by $$R^{2}$$ then allows us to relate this to the slope of the graph in Figure b [particularly the first part of the graph], which we estimate to be $$80 \mu \mathrm{V} / \mathrm{m}^{2}$$

(a) Thus, $$\dfrac{d
B_{1}}{d t}=\left(80 \mu \mathrm{V} / \mathrm{m}^{2}\right) / \pi \approx 25
\mu \mathrm{T} / \mathrm{s}$$

(b) Similar reasoning for region 2 (corresponding to the slope of the second part of the graph in Figure b) leads to an emf equal to

$$\pi
r_{1}^{2}\left(\dfrac{d B_{1}}{d t}-\dfrac{d B_{2}}{d t}\right)+\pi R^{2} \dfrac{d
B_{2}}{d t}$$

which means the second slope (which we estimate to be $$40 \mu \mathrm{V} / \mathrm{m}^{2}$$ ) is equal to $$\pi \dfrac{d B_{2}}{d t}$$

$$\text {
Therefore, } \dfrac{d B_{2}}{d t}=\left(40 \mu \mathrm{V} /
\mathrm{m}^{2}\right) / \pi \approx 13 \mu \mathrm{T} / \mathrm{s}$$

(c) Considerations of Lenz's law leads to the conclusion that $$B_{2}$$ is
increasing.