welcome back to physics 1308

Physics 1308: General Physics II - Professor Jodi Cooley

Michael Faraday 22 September 1791 – 25 August 1867

Induction and Inductance

Welcome Back to Physics 1308


Announcements

• Assignments for Tuesday, November 6th:

- Reading: Chapter 30.6 - 30.8

- Watch Videos:

- https://youtu.be/y56A0BZGuuU — Lecture 19 -Magnetic Induction (56:54 - end)

• Homework 10 Assigned - due before class on Tuesday, November 6th.

https://youtu.be/y56A0BZGuuU


Review Question 1An initially unmagnetized iron bar is placed next to a solenoid. Which one of the following statements describes the iron bar after the solenoid is connected to the battery?

A) A magnetic force accelerates the bar to the right. B) Since the bar is unmagnetized, there will not be any affect on

the bar. C) The magnetic field of the solenoid will cause a current to

flow in a loop that extends from one end of the bar to the other and that continues until the battery is disconnected from the solenoid.

D) The magnetic field of the solenoid induces magnetism in the bar with the bar’s north pole nearest to the solenoid.

E) The magnetic field of the solenoid induces magnetism in the bar with the bar’s south pole nearest to the solenoid.


Review Question 2A wire, connected to a battery and switch, passes through the center of a long current-carrying solenoid as shown in the drawing. When the switch is closed and there is a current in the wire, what happens to the portion of the wire that runs inside of the solenoid?

A) There is no effect on the wire. B) The wire is pushed downward. C) The wire is pushed upward. D) The wire is pushed toward the left. E) The wire is pushed toward the right


Review Key Concepts• Electric charge exists in two kinds, positive and negative. These exert influence on each other via

a field of force.

• The electric field is a conservative force field, with associated concepts of potential energy and the new concept of “electric potential”. Fields accelerate charges, and this is the basis of circuits. There are relationships between fields and the energy stored and release in circuits (e.g. in capacitors and resistors).

• Magnetic fields, fields of force exerted by certain materials such as magnetite and iron, also affect electric charges. However, they do so at right angles to the field lines and the direction of motion of the charges.

• Moving electric charge creates magnetic field. Changing magnetic field penetration in a conductor causes voltage, and thus electric current. A conductor that encloses its OWN magnetic field possesses “self-induction”. Magnetic fields store energy, just like electric fields.


Key ConceptsFaraday’s Law of Induction:

The induced emf tends to oppose the flux change and the minus sign indicates this opposition. This minus sign is referred to as Lenz’s Law.

Magnetic Flux:

�B = (B cos ✓)A


Key ConceptsLenz’ Law:

An induced current has a direction such that the magnetic field due to this induced current opposes the change in the magnetic flux that induces the current. The induced emf has the same direction as the induced current.

As the magnet is moved toward the loop, a current is induced in the loop. The current produces its own magnetic field, with magnetic dipole moment µ oriented so as to oppose the motion of the magnet. Thus, the induced current must be counterclockwise as shown.


Question 1A long straight wire carries a steady current I. A rectangular conducting wire loop lies in the same plane as the wire with two sides parallel to the wire and two sides perpendicular. Suppose the loop is pushed toward the wire as shown. Given the direction of I, the induced current in the loop is

A) clockwise B) counterclockwise C) no current is induced

As the loop approaches the wire, the magnetic flux through the loop will increase. To counteract the increasing magnetic flux, a current will start flowing to create a counteracting magnetic flux through the loop. Since the magnetic flux from the wire points into the page, the counteracting magnetic flux must point out of the page. This requires a CCW current.


Question 2A rigid, circular metal loop begins at rest in a uniform magnetic field directed away from you as shown. The loop is then pulled through the field toward the right, but does not exit the field. What is the direction of any induced current within the loop?

A) clockwise B) counterclockwise C) no current is induced

In this case there is no change in the magnetic flux. Hence their is no induced current.


Instructor Problem: Earring and MRI MachineYou accidentally wear a small, circular (one loop) 2.0 g gold earring when entering a room containing the field from a Magnetic Resonance Imaging (MRI) machine. The earring has a radius of 0.50 cm and the gold has a resistance of 8.8 x 10-4 Ω. The field makes an angle of 37° with respect to the vector normal to the plane of the earring. As you walk in you do so in a direction of increasing magnetic field strength, going from 0.0 T to 1.5 T in 1.5 s. A) Calculate the voltage established in the earring due to your motion through the magnetic

field. B) Calculate the electric current that now flows through the earring due to the induced

voltage, and the power dissipated by the earring. C) Calculate the magnitude of the torque exerted on the earring due to the induced magnetic

dipole moment in the earring when the external field strength reaches 1.5 T. (assume the angle given hasn't changed up to that point). What is the angular acceleration of the hoop earring about an axis perpendicular to the normal?


Given: m = 2.0 g

r = 0.50 cm

R = 8.8⇥ 10�4 ⌦

✓ = 37�

dB = (1.5� 0.0)T

dt = 1.5 s

Part A:This is a problem about a changing flux inducing a voltage in a conductor. Thus, we need Faraday’s Law.

E = �d�B

dt

�B = BA cos ✓

The magnetic flux is given by

The only thing in the fl ︎ux equation that is changing is the strength of the magnetic ︎field. Thus,

E = �A cos ✓dB

dt

= �(⇡ ⇥ (0.005 m)2) cos (37�)(1.5� 0.0)T

1.5 s

E = �6.3⇥ 10�5 V


Part B:

Now we want to find the current. Since the earring is gold, we can just use Ohm’s Law.

E = IR �! I =ER

=6.3⇥ 10�5 V

8.5⇥ 10�4 ⌦�! I = 71 mA

Part C:

Finally, we want to find the torque exerted on the earring by the MRI field once the field reaches full strength. Torque is given by

~⌧ = ~µ⇥ ~B

We know that the magnitude of the magnetic moment is given by

and the magnitude is given by⌧ = µB sin ✓

µ = NiA

Thus,⌧ = NiAB sin ✓ = (1)(0.071 A)(⇡(0.005 m)2)(1.5 T ) sin 37� �! ⌧ = 5.1⇥ 10�6 N ·m


The second part of this problem asks about the angular acceleration about an axis perpendicular to the normal. This is given by

⌧ = I↵ �! ↵ =⌧

I

So, we need the moment of inertia for a hoop.

I =1

2mr2

Putting it together…

↵ =2⌧

mr2=

2(5.1⇥ 10�6 N ·m)

(0.002 g)(0.005 m)2

↵ = 204 rad/s2


Student Problem: Magnetic Induction You are standing below a high-tension power line. This particular power line is carrying an alternating current whose maximum value is 1100 A. It decreases from that to zero in 0.0083 s. Treat the power line as a perfectly straight line of current and the current change as linear.

You are holding a thin metal ruler with an area of 0.305 m2 (12 in x 1 in), and the ruler is 65 ft below the power line. What is the maximum voltage that can be induced in the ruler by the changing current in the high-tension power line?


This problem involves a changing current creating a changing magnetic field which can penetrate a conductor, creating a changing ︎flux, which creates a voltage in the conductor according to Faraday's Law.

Given: A = 0.305 m2

D = 65 ftdt = 0.0083 s

First we need to find the magnetic field created by a straight line. From the video lecture (and the textbook we know that this is given by

B =µ0i

2⇡D

imax = 1100 A

imin = 0 A

=18.812 m

Thus,

Bmax =(1.26⇥ 10�6 T ·m/A)(1100 A)

2⇡(19.812 m)= 1.11⇥ 10�5 T

= 0.00 TBmin =(1.26⇥ 106 T ·m/A)(0.00 A)

2⇡(19.812 m)


Now we can apply Faraday’s law of induction to determine the voltage created by the changing magnetic field.

E = �d�

dt

The area of the ruler does not change — so, we want to choose θ such that the resulting voltage is as big as possible. To get the maximal flux we need

cos ✓ = 1

= �A cos ✓dB

dt

Putting this together

E = �AdB

dt= �(0.305 m)

(1.1⇥ 10�5 � 0)T

0.0083 s

E = 4.1⇥ 10�4 V Note: This is a really tiny voltage that is induced — only about 1/2 mV. Also, I drop the (-) because we really only care about magnitude. Your book does this in the examples as well.

�! ✓ = 0


The End for Today!

welcome back to physics 1308

Documents