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Significance

An alternative way of looking at the induced emf from Faraday’s law is to integrate in space instead of time. The solution, however, would be the same. The motional emf is

| ε | = B v d l .

The velocity can be written as the angular velocity times the radius and the differential length written as dr . Therefore,

| ε | = B v d r = B ω 0 l r d r = 1 2 B ω l 2 ,

which is the same solution as before.

A rectangular coil rotating in a magnetic field

A rectangular coil of area A and N turns is placed in a uniform magnetic field B = B j ^ , as shown in [link] . The coil is rotated about the z -axis through its center at a constant angular velocity ω . Obtain an expression for the induced emf in the coil.

Figure shows a rectangular coil rotating in a uniform magnetic field.
A rectangular coil rotating in a uniform magnetic field.

Strategy

According to the diagram, the angle between the perpendicular to the surface ( n ^ ) and the magnetic field ( B ) is θ . The dot product of B · n ^ simplifies to only the cos θ component of the magnetic field, namely where the magnetic field projects onto the unit area vector n ^ . The magnitude of the magnetic field and the area of the loop are fixed over time, which makes the integration simplify quickly. The induced emf is written out using Faraday’s law.

Solution

When the coil is in a position such that its normal vector n ^ makes an angle θ with the magnetic field B , the magnetic flux through a single turn of the coil is

Φ m = S B · n ^ d A = B A cos θ .

From Faraday’s law, the emf induced in the coil is

ε = N d Φ m d t = N B A sin θ d θ d t .

The constant angular velocity is ω = d θ / d t . The angle θ represents the time evolution of the angular velocity or ω t . This is changes the function to time space rather than θ . The induced emf therefore varies sinusoidally with time according to

ε = ε 0 sin ω t ,

where ε 0 = N B A ω .

Significance

If the magnetic field strength or area of the loop were also changing over time, these variables wouldn’t be able to be pulled out of the time derivative to simply the solution as shown. This example is the basis for an electric generator, as we will give a full discussion in Applications of Newton’s Law .

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Check Your Understanding Shown below is a rod of length l that is rotated counterclockwise around the axis through O by the torque due to m g . Assuming that the rod is in a uniform magnetic field B , what is the emf induced between the ends of the rod when its angular velocity is ω ? Which end of the rod is at a higher potential?

Figure shows a rod of length l that is located in the uniform magnetic field. The rod is rotated counterclockwise around the axis through O by the torque due to mg.

ε = B l 2 ω / 2 , with O at a higher potential than S

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Check Your Understanding A rod of length 10 cm moves at a speed of 10 m/s perpendicularly through a 1.5-T magnetic field. What is the potential difference between the ends of the rod?

1.5 V

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Summary

  • The relationship between an induced emf ε in a wire moving at a constant speed v through a magnetic field B is given by ε = B l v .
  • An induced emf from Faraday’s law is created from a motional emf that opposes the change in flux.

Conceptual questions

A bar magnet falls under the influence of gravity along the axis of a long copper tube. If air resistance is negligible, will there be a force to oppose the descent of the magnet? If so, will the magnet reach a terminal velocity?

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Questions & Answers

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Practice Key Terms 1

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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