91影视

1. Area of the loop$\mathrm{A}=6.8{\mathrm{mm}}^2=6.8脳{10}^{-6}{\mathrm{m}}^2$
2. Turns/cm in the solenoid$\mathrm{n}=854\frac{\mathrm{turns}}{\mathrm{cm}}=85400\frac{\mathrm{turns}}{\mathrm{m}}$
3. Current through the solenoid$\mathrm{I}=1.69脳{10}^{-8}\mathrm{A}-\mathrm{m}$
4. Angular frequency of the current f = 212 rad/s

By using the concept of the solenoid and Faraday鈥檚 law, find the amplitude of the induced emf.

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

Formulae are as follows:

$$\begin{gathered} B={渭}_{0}nl \\ 蔚=\frac{d蠁}{dt} \\ 蠁=蠒BdA \\ t=\frac{1}{f} \end{gathered}$$

A long, tightly wound helical coil of wire is called a solenoid.

When the current flows through the wire, the magnetic field induced inside the solenoid is given by,

$$\begin{gathered} \mathrm{B}={\mathrm{渭}}_0\mathrm{nl} \\ \mathrm{B}=4脳\mathrm{蟿蟿}脳{10}^{-7}脳85400脳1.28 \\ \mathrm{B}=0.1372\mathrm{T}.......................................................................\left(1\right) \end{gathered}$$

By using the frequency of the current, find the time with which the flux is changing through the loop.

$$\begin{gathered} t=\frac{1}{f} \\ \mathrm{t}=\frac{1}{212}.......................................................................\left(2\right) \end{gathered}$$

By Faraday鈥檚 law,

$$\begin{gathered} 蔚=\frac{d蠁}{dt} \\ 蠁=蠒BdA=BA \\ 蔚=\frac{d\left(BA\right)}{dt} \\ 蔚=\frac{AB}{t} \end{gathered}$$

Using the value of A in equation 1 and 2,

$蔚=\frac{6.8脳{10}^{-6}脳0.1372}{{\displaystyle \frac{1}{212}}}$

$$\begin{gathered} \mathrm{蔚}=6.8脳{10}^{-6}脳0.1372脳212 \\ \mathrm{蔚}=198.50脳{10}^{-6} \\ \mathrm{蔚}=0.198脳{10}^{-3}\mathrm{V} \\ \mathrm{蔚}=0.198\mathrm{mV} \end{gathered}$$

Hence, amplitude of the emf induced in the loop is,$\mathrm{蔚}=0.198\mathrm{mV}$.

Therefore, by using the concept of the solenoid and Faraday鈥檚 law, the amplitude of the induced emf can be determined.