Question: In Figure, a stiff wire bent into a semicircle of radius a = 2.0cmis rotated at constant angular speed $\mathbf{40}\frac{\mathbf{rev}}{\mathbf{s}}$in a uniform 20mTmagnetic field. (a) What is the frequency? (b) What is the amplitude of the emf induced in the loop?

1. Magnetic field$B=20\mathrm{mT}=20.0\times {10}^{-3}\mathrm{T}$
2. Radius of semicircle$a=2.0cm=20.0\times {10}^{-2}\mathrm{T}$
3. Angular Speed$\omega =40\mathrm{rev}/\mathrm{s}=251.32\mathrm{rad}/\mathrm{s}$

The angular speed and frequency is related. From Faraday’s law, evaluate the induced emf from 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:

$$\begin{gathered} \omega =2\mathrm{\pi f} \\ \mathbf{\varphi }=\int B.dA \\ \epsilon =-N\frac{d\mathbf{\varphi }}{dt} \end{gathered}$$

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

The angular speed is,

$\omega =\frac{40\mathrm{rev}}{s}=40\frac{\mathrm{rev}}{\mathrm{s}}\times 2\pi \frac{\mathrm{rad}}{\mathrm{rev}}$

The frequency and the angular speed is related as,

$\omega =2\mathrm{\pi }f$

Where,is frequency in,

$\omega =\frac{40\mathrm{rev}}{s}\times 2\pi \frac{\mathrm{rad}}{\mathrm{rev}}=2\mathrm{\pi f}$

Thus,

f = 40 Hz

Hence, frequency is 40.0Hz

The total magnetic lines are,

$\mathbf{\varphi }=\int \overline{B}.\overline{d}\overline{A}=BA\cos \theta =BA\cos \left(\psi t\right)$

From Faraday’s law,

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

For the semicircle area,$A=\frac{1}{2}\left(\pi r^2\right)$

$$\begin{gathered} \epsilon =\left(\pi r^2\right)B\omega \sin \left(\omega t\right) \\ \epsilon =\frac{1}{2}\left(\pi \times 0.{02}^2\right)\times \left(20.0\times {10}^{-3}T\right)\times 251.3\mathrm{rad}/\mathrm{s}\times \sin \left(90°\right) \\ \mathrm{\epsilon }=3.158\times {10}^{-3}\mathrm{V}\approx 3.2\mathrm{mV} \end{gathered}$$

Hence, amplitude of the emf induced in the loop is,$\mathrm{\epsilon }=3.2\mathrm{mV}\approx 3.{210}^{-3}\mathrm{V}$.

Therefore, the frequency can be determined from the angular speed. The amplitude of the induced emf can be found using Faraday’s law for the shown figure in which thewire is bent into a semicircle.