Question

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?

Short answer

1. Frequency is 40.0Hz
2. Amplitude of the emf induced in the loop is,$\epsilon =3.2\mathrm{mV}=3.2\times {10}^{-3}\mathrm{V}$.

Step-by-step solution

Step 1: Given

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$\omega =40\mathrm{rev}/\mathrm{s}=251.32\mathrm{rad}/\mathrm{s}$

Determining the concept

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.

(a) Determining the 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

(b) Determining the amplitude of the induced emf in the loop

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.