In Figure, a 120 turncoil of radius 1.8 cm and resistance $\mathbf{5}\mathbf{.}\mathbf{3}\mathbf{\Omega }$is coaxial with a solenoid of 220 turne/cm and diameter 3.2 cm. The solenoid current drops from 1.5 Ato zero in time interval $\mathbf{∆}\mathit{t}\mathbf{=}\mathbf{25}\mathit{m}\mathit{s}$. What current is induced in the coil during$\mathbf{∆}\mathit{t}$?

1. The total number of turns of the coil is, N = 120
2. The radius of the coil is, 1.8 cm
3. The resistance is,$\mathrm{R}=5.3\mathrm{\Omega }$
4. The number of turns per meter is, n=220 turns/m= 22000 turns/m
5. The diameter of the solenoid is, 3.2 cm
6. The current drop from 1.5 A to 0 in time interval $∆t=25\mathrm{ms}$.

Find the emf induced due to the change in the magnetic flux. Now, applying Ohm’s law, find the current induced in the coil during $\mathbf{∆}\mathit{t}$.

Faraday'slaw 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:

$$\begin{gathered} {\Phi }_B=BA \\ B={\mu }_{0}in \\ \epsilon =-N\frac{d{\Phi }_B}{dt} \\ i=\frac{\left|\epsilon \right|}{R} \end{gathered}$$

Where,${\Phi }_B$is magnetic flux, B is magnetic field, A is area, i is current, R is resistance,𝜀 is emf, Nis number of turns,𝜇₀is permeability.

From Eq.30-2, the magnetic flux is ,

${\Phi }_B=BA...................(30-2)$

Also, from Eq.29-23, the magnitude of the magnetic field is given by,

$B={\mu }_{0}in.......................(29-23)$

According to the Faraday’s law, emf is induced due to the change in the magnetic flux is,

$\epsilon =-N\frac{d{\Phi }_B}{dt}.......................(30-5)$

Substituting Eq.30-2 in Eq.30-5,

$\epsilon =-NA\frac{d\left(B\right)}{dt}$

Now, substituting Eq.29-23 in Eq.30-5,
$$\begin{gathered} \epsilon =-NA\frac{d\left({\mu }_{0}in\right)}{dt} \\ \epsilon =-Nn{\mu }_{0}A\frac{di}{dt} \end{gathered}$$

But, putting $A=\tau \tau r^2$,

$\epsilon =-Nn{\mu }_0\left(\tau \tau r^2\right)\frac{di}{dt}$

Substituting given values,

$$\begin{gathered} \epsilon =-\left(120\right)\left(22000\frac{\mathrm{turns}}{\mathrm{m}}\right)\left(4\mathrm{\tau \tau }\times {10}^{-7}\mathrm{T}.\frac{\mathrm{m}}{\mathrm{A}}\right)\left[\mathrm{\tau \tau }{(0.016\mathrm{m})}^2\right]\left(\frac{1.5\mathrm{A}}{0.025\mathrm{s}}\right) \\ \mathrm{\epsilon }=-0.16\mathrm{V} \end{gathered}$$

According to Ohm’s law,

$$\begin{gathered} i=\frac{\left|\epsilon \right|}{R} \\ i=\frac{0.16V}{5.3\Omega } \\ i=0.030A \end{gathered}$$

Hence, the current induced in the coil during $∆t$is $i=0.030A$.

Therefore, by using the concept of Faraday’s law and Ohm’s law, the current induced in the coil can be determined.