Solve

Guides

0

Question

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.

Open in App

Solution

Verified by Toppr

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 $$a=0.0080 \mathrm{T} / \mathrm{s}$$

Was this answer helpful?

3