Figure (a) shows a circuit consisting of an ideal battery with emf $\mathit{\epsilon }\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{}\mathit{m}\mathit{V}$, a resistance R, and a small wire loop of area $\mathbf{500}\mathbf{}\mathit{c}{\mathit{m}}^{\mathbf{2}}$. 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 Tesla, 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${\mathit{i}}_{\mathbf{s}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}\mathit{A}$. Find the constant a in the equation for the field magnitude.

i) Emf of the battery is,${\epsilon }_{battery}=6.00\mu V$

ii) Wire loop area,$A=5.0c{m}^2$

iii) Time interval t = 10 s to t = 20 s

iv) Fig.30-37

v) The field magnitude is B = at

vi) The vertical axis scale is set by$i_s=2.0mA$

vii) The horizontal axis scale is set by$t_s=30s$ .

Applying Ohm’s law, find the value of resistance and using this value offind the value of inducedemf during$\mathbf{10}\mathit{s}\mathbf{<}\mathit{t}\mathbf{<}\mathbf{20}\mathit{s}$. Using Eq.30-2 in Eq.30-4, find theconstantin the equation for the field magnitude.

Faraday's law of electromagnetic inductionstates, Whenever a conductor is placed in a varying magnetic field, an electromotive force is induced in it.

Ohm's law states that the voltage across a conductor is directly proportional to the current flowing through it, provided all physical conditions and temperatures remain constant.

Formulae are as follow:

$$\begin{gathered} i=\frac{{\epsilon }_{battery}}{R} \\ {\epsilon }_{induced}=iR-{\epsilon }_{battery} \\ {\Phi }_B=BA \\ {\epsilon }_{induced}=-\frac{d{\Phi }_B}{dt} \end{gathered}$$

Where,${\Phi }_B$is magnetic flux, B is magnetic field, A is area, i is current, R is resistance, 𝜀 is emf.

From Fig.30-37(b), at t = 0 , i = 0.0015A .

Applying Ohm’s law, the current i is given by,

$i=\frac{{\epsilon }_{battery}}{R}$

Therefore, the resistance is given by,

$$\begin{gathered} R=\frac{6.00\times {10}^{-6}V}{0.0015A} \\ R=0.0040\Omega \end{gathered}$$

Now, during$10s<t<20s$ , the value of the current is ,

$i=\frac{\left({\epsilon }_{battery}+{\epsilon }_{induced}\right)}{R}$

Therefore, the induced emf is,

${\epsilon }_{induced}=iR-{\epsilon }_{battery}$

From Fig.30-37(b) , i = 0.0050A

$$\begin{gathered} {\epsilon }_{induced}=\left(0.00050A\right)\left(0.0040\Omega \right)-\left(6.00\times {10}^{-6}V\right) \\ {\epsilon }_{induced}=-4.00\times {10}^{-6}V \end{gathered}$$

Now, from Eq.30-2, magnetic flux is,

${\varphi }_B=BA................................................................\left(30-2\right)$

According to Faraday’s law, the induced emf is,

${\epsilon }_{induced}=-\frac{d{\Phi }_B}{dt}..............................................................\left(30-4\right)$

Putting Eq.30-2,

$$\begin{gathered} {\epsilon }_{induced}=-\frac{d\left(BA\right)}{dt} \\ {\epsilon }_{induced}=-A\frac{dB}{dt} \end{gathered}$$

But,

$$\begin{gathered} \frac{dB}{dt}=a \\ {\epsilon }_{induced}=-Aa \end{gathered}$$

Therefore, the value of is,

$$\begin{gathered} a=-\frac{{\epsilon }_{induced}}{A} \\ a=-\frac{\left(-4.00\times {10}^{-6}V\right)}{\left(5.0\times {10}^{-4}{m}^2\right)} \\ a=0.0080T/s \end{gathered}$$

Hence, the value of constant a is 0.0080 T/s .

Therefore, by using the concept of Faraday’s law and Ohm’s law, the value of constant a can be deteremined.