Figure a shows a circuit consisting of an ideal battery with emf $$\mathscr{E}=6.00\muV$$, a resistance R, and a small wire loop of area $$5.0cm^2$$. For the time interval t = 10 s to t = 20 s, an external magnetic field is set up throughout the loop.The field is uniform, its direction is into the page in Figure a, and the field magnitude is given by $$B=at$$, where B is in teslas, a is a constant, and t is in seconds. Figure b gives the current $$i$$ in the circuit before, during, and after the external field is set up. The vertical axis scale is set by $$i_s=2.0mA$$. Find the constant a in the equation for the field magnitude.
From the datum at $$t=0$$ in Fig. b
we see $$0.0015 \mathrm{A}=V_{\text {battery }} / R,$$ which implies that the
resistance is
$$R=(6.00 \mu \mathrm{V}) /(0.0015
\mathrm{A})=0.0040 \Omega$$
Now, the value of the current during
$$10 \mathrm{s}<t<20 \mathrm{s}$$ leads us to equate
$$\left(V_{\text {battery
}}+\varepsilon_{\text {induced }}\right) / R=0.00050 \mathrm{A}$$
This shows that the induced emf is $$\varepsilon_{\text
{induced }}=-4.0 \mu \mathrm{V}$$. Now we use Faraday's law:
$$\varepsilon=-\dfrac{d \Phi_{B}}{d
t}=-A \dfrac{d B}{d t}=-A a$$
Plugging in $$\varepsilon=-4.0
\times 10^{-6} \mathrm{V}$$ and $$A=5.0 \times 10^{-4} \mathrm{m}^{2},$$ we
obtain $$a=0.0080 \mathrm{T} / \mathrm{s}$$