Question

A certain elastic conducting material is stretched into a circular loop of 12.0 cm radius. It is placed with its plane perpendicular to a uniform 0.800 Tmagnetic field. When released, the radius of the loop starts to shrink at an instantaneous rate of 75.0cm/s. What emf is induced in the loop at that instant?

Short answer

The magnetic flux through the loop is $\epsilon =0.452\mathrm{V}$.

Step-by-step solution

Given

1. The radius of the loop is, r = 12.0 cm =0.120 m
2. The plan of elastic conductor is perpendicular to the uniform magnetic field.
3. the uniform magnetic field is, B = 0.800 T
4. the loop is starts to shrink at an instantaneous rate of$\frac{\mathrm{dr}}{\mathrm{dt}}=-75.0\frac{\mathrm{cm}}{\mathrm{s}}=0.750\mathrm{m}/\mathrm{s}$

Determining the concept

Using the equation for the magnetic flux and the given information and applying Faraday’s law, find the emf induced in the loop.

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:

The magnetic flux through the loop is,

$$\begin{gathered} {\Phi }_B=BA\cos \theta \\ \epsilon =-\frac{d{\Phi }_B}{dt} \end{gathered}$$

Where,${\Phi }_B$is magnetic flux, B is magnetic field, A is area, 𝜀 is emf.

Determining the magnetic flux through the loop

From Eq.30-2, the magnetic flux through the loop is,

${\Phi }_B=BA\cos \theta ...............(30-2)$

where,$\cos \theta =1$,since the magnetic field is perpendicular to theplane of the elastic conductor.

According to Faraday’s law, the emf induced due to the change in the magnetic flux is,

$\epsilon =-\frac{d{\Phi }_B}{dt}..............(30-4)$

Form Eq.30-2,

$\epsilon =-\frac{d\left(BA\right)}{dt}$

But the magnetic field is uniform and the area is, $A=\tau \tau r^2$. Thus,

$$\begin{gathered} \epsilon =-B\frac{d\left(\tau \tau r^2\right)}{dt} \\ \epsilon =-2\mathrm{\tau \tau rB}\frac{\mathrm{dr}}{\mathrm{dt}} \\ \epsilon =-2\mathrm{\tau \tau }(0.120\mathrm{m})(0.800\mathrm{T})(-0.750\mathrm{m}/\mathrm{s}) \\ \epsilon =0.452\mathrm{V} \end{gathered}$$

Hence, the magnetic flux through the loop is$\epsilon =0.452V$.

Therefore, by using Faraday’s law and equation, the magnetic flux through the loop can be determined.