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

Expert verified

The magnetic flux through the loop is ε=0.452V.

Step by step solution

01

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 ofdrdt=-75.0cms=0.750m/s
02

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,

ΦB=BAcosθε=-dΦBdt

Where,ΦBis magnetic flux, B is magnetic field, A is area, 𝜀 is emf.

03

Determining the magnetic flux through the loop

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

ΦB=BAcosθ...............(30-2)

where,cosθ=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,

ε=-dΦBdt..............(30-4)

Form Eq.30-2,

ε=-d(BA)dt

But the magnetic field is uniform and the area is, A=ττr2. Thus,

ε=-Bd(ττr2)dtε=-2ττrBdrdtε=-2ττ(0.120m)(0.800T)(-0.750m/s)ε=0.452V

Hence, the magnetic flux through the loop isε=0.452V.

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

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