One hundred turns of (insulated) copper wire are wrapped around a wooden cylindrical core of cross-sectional area $\mathbf{1}\mathbf{.}\mathbf{20}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}{\mathbf{m}}^{\mathbf{2}}$. The two ends of the wire are connected to a resistor. The total resistance in the circuit is$\mathbf{13}\mathbf{.}\mathbf{0}\mathbf{\Omega }$. If an externally applied uniform longitudinal magnetic field in the core changes from 1.60 Tin one direction to1.60 T in the opposite direction, how much charge flows through a point in the circuit during the change?

1. Cross-sectional area,$\mathrm{A}=1.20\times {10}^{-3}{\mathrm{m}}^2$
2. Resistance, $R=13.0\Omega$
3. Magnetic field at core, B (0) = 1.60 T
4. Magnetic field at other end in opposite direction, B (t) = (-1.60 T)

The magnetic flux ${\mathit{\phi }}_{\mathit{B}}$through area Ain a magnetic field is given by Faraday’s law of induction. If the magnetic field$\overrightarrow{\mathbf{B}}$is perpendicular to the area A, thensubstitute the given values in the formula to find the charge.

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

Formulae are as follows:

$emf=-N.\frac{d\phi }{dt}$

Where,$d\Phi$ is magnetic flux, N is number of turns, dt is time.

According to Faraday’s law,

$emf=-N.\frac{d\phi }{dt}$

From Ohm’s law, emf = i.R and i = dq/dt

$\therefore i.R=-N.\frac{d\phi }{dt}$

$\frac{dq}{dt}=-N.\frac{d\phi }{dt}$

Integrating this with respect to time,

$q{\left(t\right)}_0^t=-N\phi {\left(t\right)}_0^t$

To calculate the charge flowing through the point,

$q\left(t\right)=\frac{N}{R}\left[{\phi }_B\left(0\right)-{\phi }_B\left(t\right)\right]$

According to Faraday’s law, if the magnetic field $\overrightarrow{B}$is perpendicular to the area A, then,

${\phi }_B=BA$

Therefore,

$$\begin{gathered} \mathrm{q}\left(\mathrm{t}\right)=\frac{\mathrm{NA}}{\mathrm{R}}\left[\mathrm{B}\left(0\right)-\mathrm{B}\left(\mathrm{t}\right)\right] \\ \mathrm{q}\left(\mathrm{t}\right)=\frac{100\times \left(1.20\times {10}^{-3}{\mathrm{m}}^2\right)}{\left(13.0\mathrm{\Omega }\right)}\left[16.0\mathrm{T}-\left(-1.60\mathrm{T}\right)\right] \\ \mathrm{q}\left(\mathrm{t}\right)=100\times \left(0.0923\times {10}^{-3}{\mathrm{m}}^2\right)\times \left(3.20\right) \\ \mathrm{q}\left(\mathrm{t}\right)=2.95\times {10}^{-2}\mathrm{C} \end{gathered}$$

Hence, the charge flowing through a point in the circuit during the change in magnetic field is,$q\left(t\right)=2.95\times {10}^{-2}\mathrm{C}$

Therefore, the charge flow through a point in the circuit during the change in the magnetic field can be found by using Faraday’s law.