Question: An electric generator contains a coil of 100 turnsof wire, each forming a rectangular loop 50.0cm to 30.0cm. The coil is placed entirely in a uniform magnetic field with magnitude B = 3.50 Tand with$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{B}}$initially perpendicular to the coil’s plane. What is the maximum value of the emf produced when the coil is spun at 1000 rev/min about an axis perpendicular to$\stackrel{\rightharpoonup }{\mathit{B}}$?

1. Turns of wire N = 100
2. Rectangular loop dimension 50.0 cm by 30.0 cm
3. Magnetic field B = 3.50 T
4. Coil rotation 1000 rev/min

From Faraday’s law, evaluate the induced emf from the number of turns and the change in the magnetic field lines that pass through the loop. Find the magnetic field lines with the given magnetic field and the area of the rectangular loop.

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

Formulae are as follows:

$$\begin{gathered} \mathbf{\varphi }=\int \mathrm{B}.\mathrm{dA} \\ \mathrm{\epsilon }=-\mathrm{N}\frac{\mathrm{d}\mathbf{\varphi }}{\mathrm{dt}} \end{gathered}$$

Where,$\mathbf{\varphi }$ is magnetic flux, B is magnetic field, A is area, 𝜀 is emf, Nis number of turns.

The coil rotation is,

$$\begin{gathered} 1000\mathrm{rev}/\min =\frac{1000\mathrm{rev}}{1\min }\times \frac{1\min }{60\mathrm{s}}\times \frac{2\mathrm{\pi }\mathrm{rad}}{1\mathrm{rev}}=104.7\mathrm{rad}/\sec \\ \mathrm{\omega }=104.7\mathrm{rad}/\sec \end{gathered}$$

The magnetic field lines $\mathbf{\varphi }$with the given magnetic field and the area of the rectangular loop is given as,

$$\begin{gathered} \mathbf{\varphi }=\int B.dA \\ \mathbf{\varphi }=\int BdA\cos \theta \end{gathered}$$

But$\theta =\omega trad$

$$\begin{gathered} \mathbf{\varphi }=\int BdA\cos \left(\omega t\right) \\ \mathbf{\varphi }=BdA\cos \left(\omega t\right) \end{gathered}$$

By differentiating the magnetic field lines with respect to time,

$$\begin{gathered} \frac{\mathbf{d}\mathbf{\varphi }}{\mathbf{dt}}=\frac{d}{dt}BA\cos \omega t \\ \frac{\mathbf{d}\mathbf{\varphi }}{\mathbf{dt}}=-\omega BA\cos \left(\omega t\right) \end{gathered}$$

From Faraday’s law, induced emf,

$$\begin{gathered} \epsilon =-N\frac{\mathrm{d}\mathbf{\varphi }}{\mathrm{dt}} \\ \epsilon =-N\left(-\omega BA\sin \left(\omega t\right)\right)=N\omega BA\sin \\ \epsilon =100\times 104.7\times 3.5\times \left(0.5\times 0.3\right)\sin \left(\omega t\right) \end{gathered}$$

The induced emf will be maximum, when $\sin \left(\omega t\right)=1$.

$$\begin{gathered} \epsilon =100\times \left(104.7\mathrm{rad}/\mathrm{s}\right)\times \left(3.5\mathrm{T}\right)\times \left(0.5m\times 0.3m\right)\times 1 \\ \epsilon =5.496\times {10}^3\gg 5.5\mathrm{kV} \end{gathered}$$

Hence, the maximum value of emf is, $\epsilon =5.5\mathrm{kV}$.

Therefore, the induced emf in an electric generator can be determined using Faraday’s law when a coil of wire is wounded with the given number of turns.