Question

A small loop of area 6.8 ${\mathbf{mm}}^{\mathbf{2}}$is placed inside a long solenoid that hasand carries a sinusoidally varying current i of amplitude1.28 A and angular frequency rad/s.The central axes of the loop and solenoid coincide. What is the amplitude of the emf induced in the loop?

Short answer

Amplitude of the emf induced in the loop is$\mathrm{\epsilon }=0.198\mathrm{mV}$

Step-by-step solution

Given

1. Area of the loop$\mathrm{A}=6.8{\mathrm{mm}}^2=6.8\times {10}^{-6}{\mathrm{m}}^2$
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$\mathrm{I}=1.69\times {10}^{-8}\mathrm{A}-\mathrm{m}$
4. Angular frequency of the current f = 212 rad/s

Determining the concept

By using the concept of the solenoid and Faraday’s 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:

$$\begin{gathered} B={\mu }_{0}nl \\ \epsilon =\frac{d\phi }{dt} \\ \phi =\varphi BdA \\ t=\frac{1}{f} \end{gathered}$$

Where, $\mathbf{\Phi }$is magnetic flux, B is magnetic field, A is area, lis current, 𝜀 is emf, nis number of turns, 𝜇₀is permeability, t is time, f is frequency.

Determining the amplitude of the emf induced in the loop

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,

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

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’s law,

$$\begin{gathered} \epsilon =\frac{d\phi }{dt} \\ \phi =\varphi BdA=BA \\ \epsilon =\frac{d\left(BA\right)}{dt} \\ \epsilon =\frac{AB}{t} \end{gathered}$$

Using the value of A in equation 1 and 2,

$\epsilon =\frac{6.8\times {10}^{-6}\times 0.1372}{{\displaystyle \frac{1}{212}}}$

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

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

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