91影视

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

The angular speed and frequency is related. From Faraday鈥檚 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:

$$\begin{gathered} 蝇=2\mathrm{蟺蹿} \\ \mathbf{蠒}=鈭�B.dA \\ 蔚=-N\frac{d\mathbf{蠒}}{dt} \end{gathered}$$

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

The angular speed is,

$蝇=\frac{40\mathrm{rev}}{s}=40\frac{\mathrm{rev}}{\mathrm{s}}脳2蟺\frac{\mathrm{rad}}{\mathrm{rev}}$

The frequency and the angular speed is related as,

$蝇=2\mathrm{蟺}f$

Where,is frequency in,

$蝇=\frac{40\mathrm{rev}}{s}脳2蟺\frac{\mathrm{rad}}{\mathrm{rev}}=2\mathrm{蟺蹿}$

Thus,

f = 40 Hz

Hence, frequency is 40.0Hz

The total magnetic lines are,

$\mathbf{蠒}=鈭�\stackrel{炉}{B}.\stackrel{炉}{d}\stackrel{炉}{A}=BA\cos 胃=BA\cos \left(蠄t\right)$

From Faraday鈥檚 law,

$$\begin{gathered} 蔚=-\frac{\mathbf{d}\mathbf{蠒}}{\mathbf{dt}} \\ 蔚=-\frac{\mathrm{d}}{\mathrm{dt}}BA\cos \left(蝇t\right) \\ 蔚=-BA\frac{\mathrm{d}}{\mathrm{dt}}\cos \left(蝇t\right) \\ 蔚=BA蝇\sin \left(蝇t\right) \end{gathered}$$

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

$$\begin{gathered} 蔚=\left(蟺r^2\right)B蝇\sin \left(蝇t\right) \\ 蔚=\frac{1}{2}\left(蟺脳0.{02}^2\right)脳\left(20.0脳{10}^{-3}T\right)脳251.3\mathrm{rad}/\mathrm{s}脳\sin \left(90掳\right) \\ \mathrm{蔚}=3.158脳{10}^{-3}\mathrm{V}鈮�3.2\mathrm{mV} \end{gathered}$$

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

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