Chapter 29 electromagnetic induction SV

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Magnetic flux:
Faraday’s Law of Induction:

The induced emf in a closed loop equals the negative of the time
rate of change of the magnetic flux through the loop

Chapter 29: ELECTROMAGNETIC INDUCTION
Direction : curl fingers of right hand around area vector A

Exercises: 1, 3, 5, 9, 17, 19, 23, 25, 27, 29, 35, 37, 39
Problems: 43(44), 45, 49(49), 55(55), 61(61), 63(63), 65(65), 67(67), 69(70), 71(72)
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Email:

N = number of turns

Coil:

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Lenz’s Law

Motional Electromotive Force
If a conductor moves in a magnetic field, a motional emf is induced.

The direction of any magnetic induction effect is such as to oppose the cause of the effect
- If the flux in a stationary circuit changes, the induced current sets up a magnetic field
opposite to the original field if original B increases, but in the same direction as original B
if B decreases.
- The induced current opposes the change in the flux through a circuit

Magnetic force FB that causes free charges in rod to move, creating excess
charges at opposite ends.
The excess charges generate an electric field (from a to b) and electric
force (F = q E) opposite to magnetic force.

Induced Electric Fields
An induced emf occurs when there is a changing magnetic flux through a stationary conductor

Charges are in equilibrium

A current (I) in solenoid sets up B along its axis, the magnetic flux is:

General form

Induced current in loop (I’): I’= ε / R

Closed conducting loop

The force that makes the charges move around the loop is not a magnetic force. There is an induced electric field in
the conductor caused by a changing magnetic flux
Induced current:

a time-varying B induces E in
stationary conductor and emf.
E is induced even when there
is no conductor.

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29.43(44). A Changing Magnetic Field. You are testing a new data-acquisition system. This system allows you to record a graph of the current
in a circuit as a function of time. As part of the test, you are using a circuit made up of a 4.00-cm-radius, 500-turn coil of copper wire
connected in series to a 600- resistor. Copper has resistivity 1.72xl0-8 .m, and the wire used for the coil has diameter 0.0300 mm. You
place the coil on a table that is tilted 30.0° from the horizontal and that lies between the poles of an electromagnet. The electromagnet
generates a vertically upward magnetic field that is zero for t < 0, equal to (0.120 T)(l - cost) for 0 < t < 1.00 s, and equal to 0.240 T for t >
1.00 s. (a) Draw the graph that should be produced by your data-acquisition system. (This is a full-featured system, so the graph will include
labels and numerical values on its axes.) (b) If you were looking vertically downward at the coil, would the current be flowing clockwise or
counterclockwise?

Displacement Current and Maxwell’s Equations
Conduction current : iC into one plate and out of the other
Displacement current (iD): fictitious current in region between capacitor’s plates.

Generalized Ampere’s Law:
Displacement current creates B between plates of capacitor while it charges

Use Faraday’s law to calculate the magnitude of the induced emf and Lenz’s law to determine its direction; Ohm’s law to calculate I

Maxwell’s Equations of Electromagnetism

d
d
dB
  NBA cos    N  r 
cos30
dt
dt
dt
For t < 0 and t > 1.00 s, dB/dt = 0   = 0  I =  /R =0


Ampere’s law

2

B

0




600 

3
For 0  t  1.00 s, dB/dt = (0.120 T) πsin (πt)    N  r    0.120 sin t 
2


i


R

  i  0.224sin t  mA 
N2r 
B     
L

R    600  
 600 
 the current is clockwise
A
r

2

Faraday’s law

2

0

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29.49(49). In Fig. the loop is being pulled to the right at constant speed v. A constant current I flows in the long wire, in
the direction shown. (a) Calculate the magnitude of the net emf  induced in the loop. Do this two ways: (i) by using
Faraday’s law of induction and (ii) by looking at the emf induced in each segment of the loop due to its motion. (b) Find
the direction (clockwise or counterclockwise) of the current induced in the loop. Do this two ways: (i) using Lenz’s law
and (ii) using the magnetic force on charges in the loop. (c) Check your answer for the emf in part (a) in the following
special cases to see whether it is physically reasonable: (i) The loop is stationary; (ii) the loop is very thin, so a  0; (iii)
the loop gets very far from the wire.

29.45. A circular coil of wire had a radius of 0.500 m, 20 turns, and a total resistance of 1.57 . The coil lies in the xy- plane. The coil is in
a uniform magnetic field B that is in the -z-direction, which is directed away from you as you view the coil. The magnitude B of the field
depends on time as follows: it increases at a constant rate from 0 at t = 0 to 0.800 T at t = 0.500 s; is constant at 0.800 T from t = 0.500 s
to t = 1.00 s; decreases at a constant rate from 0.800 T at t = 1.00 s to 0 at t = 2.00 s. (a) Graph B versus t for t from 0 to 2.00 s. (b) Graph
the current I induced in the coil versus t for t from 0 to 2.00 s. Let counterclockwise currents be positive and clockwise currents be
negative. (c) What is the maximum induced electric field magnitude in coil during 0- to 2.00-s time interval?



d
d
  NBA cos    N  r
dt
dt

B

 N
R  1.57  i    r
R R

2



2



dB
dt

(a) Consider a narrow strip of width dx and a distance x from the long wire
The magnetic field of the wire at the strip is

dB
dt

The flux through the strip is



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B

0

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d
 Iabv

dt
2r(r  a)
B

0

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For a bar of length L moving at speed v perpendicular to a magnetic field B:  = BvL
I
I
 0
bv
2  0  
 
bv
2  r  a 
2 r
Both emfs 1and 3 are directed toward the top of the loop so oppose each other. The net emf is

(c) Check your answer for the emf in part (a)

0

0

(i) The loop is stationary

4

3

1

v = 0 the induced emf should be zero

1 
 Ibv  1
 Iabv
   
 

2  r  r  a   2r(r  a)
0

1

(b) Find the direction

(ii) the loop is very thin, so a  0

0

3

When a→ 0 the flux goes to zero and the emf should approach zero
(iii) the loop gets very far from the wire

(i) Lenz’s law: The flux of the induced current opposes the change in flux

 the current is clockwise

When r →∞ the magnetic field through the loop goes to zero and the emf should go to zero

(ii) using the magnetic force: right hand ruler
segment 1
B1
B1 > B3

 FB1 > FB3

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segment 3
B3

 the induced current in the loop is clockwise (agree with Lenz’s law)

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29.61(61). The long, straight wire shown in Fig. below carries constant current I. A metal bar with
length L is moving at constant velocity v, as shown in the figure. Point a is a distance d from the
wire. (a) Calculate the emf induced in the bar. (b) Which point, a or b, is at higher potential? (c) If
the bar is replaced by a rectangular wire loop of resistance R, what is the magnitude of the current
induced in the loop?

29.55(55). Terminal Speed. A conducting rod with length L, mass m, and resistance R moves without friction
on metal rails as shown in Fig. below. A uniform magnetic field B is directed into the plane of the figure.
The rod starts from rest and is acted on by a constant force F directed to the right. The rails are infinitely
long and have negligible resistance. (a) Graph the speed of the rod as a function of time. (b) Find an
expression for the terminal speed (the speed when the acceleration of the rod is zero).

B

The magnetic field of the wire is given by


F

and varies along the length of the bar.

Divide the bar up into thin slices

Vba is negative, point a is at higher potential than point b

As the loop moves to the right the magnetic flux through it doesn’t change
The terminal speed vt occurs when the pulling force is equaled by the magnetic force

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29.63(63). A slender rod, 0.240 m long, rotates with an angular speed of 8.80 rad/s about an axis through one end and perpendicular to the
rod. The plane of rotation of the rod is perpendicular to a uniform magnetic field with a magnitude of 0.650 T. (a) What is the induced
emf in the rod? (b) What is the potential difference between its ends? (c) Suppose instead the rod rotates at 8.80 rad/s about an axis
through its center and perpendicular to the rod. In this case, what is the potential difference between the ends of the rod? Between the

center of the rod and one end?

The emf induced in a thin slice is
The emf between the center of the rod and each end is
the direction of the emf: from the center of the rod toward each end
The emfs in each half of the rod thus oppose each other: there is no net emf between the ends of the rod
Other method
2

d


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L
d
2
    t  d  dt

d  BdA  B.

2

 d  B.

L
d 1
dt   
 BL  0.165 V
2

dt
2
2

0

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29.65(65). A rectangular loop with width L and a slide wire with mass m are as shown in Fig.
below. A uniform magnetic field B is directed perpendicular to the plane of the loop into the plane
of the figure. The slide wire is given an initial speed of v0 and then released. There is no friction
between the slide wire and the loop, and the resistance of the loop is negligible in comparison to
the resistance R of the slide wire. (a) Obtain an expression for F, the magnitude of the force exerted
on the wire while it is moving at speed v. (b) Show that the distance x that the wire moves before

coming to rest is x = mv0R/a2B2

29.67(67). The magnetic field B, at all points within a circular region of radius R, is uniform in space and directed into the
plane of the page as shown in Fig. (The region could be a cross section inside the windings of a long, straight solenoid.) If
the magnetic field is increasing at a rate dB/dt, what are the magnitude and direction of the force on a stationary positive
point charge q located at points a, b, and c? (Point a is a distance r above the center of the region, point b is a distance r to
the right of the center, and point c is at the center of the region.

 the induced electric field E

(a) Use Faraday’s law to calculate the induced emf, Ohm’s law to calculate I  Lenz’s law F
The induced emf:  = Bva
the induced current

B inside; dl is clockwise  E is tangent to the circle in the counterclockwise direction

Lenz’s law:

(b) Show that the distance x
Take +x to be toward the right and let the origin be at the location of the wire at t = 0, so x 0 = 0

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29.71(72). A capacitor has two parallel plates with area A separated by a distance d. The space between plates is filled with a material
having dielectric constant K. The material is not a perfect insulator but has resistivity . The capacitor is initially charged with charge
of magnitude Q0 on each plate that gradually discharges by conduction through the dielectric. (a) Calculate the conduction current
density jC(t) in the dielectric. (b) Show that at any instant the displacement current density in the dielectric is equal in magnitude to the
conduction current density but opposite in direction, so the total current density is zero at every instant.

29.69(70). Falling Square Loop. A vertically oriented, square loop of copper wire falls from a region where the field B is horizontal,
uniform, and perpendicular to the plane of the loop, into a region where the field is zero. The loop is released from rest and initially is
entirely within the magnetic-field region. Let the side length of the loop be s and let the diameter of the wire be d. The resistivity of
copper is R and the density of copper is m. If the loop reaches its terminal speed while its upper segment is still in the magnetic-field
region, find an expression for the terminal speed.

At the terminal speed, the upward force FB exerted on the loop due to the induced
current equals the downward force of gravity: FB = mg


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









mg

FB

+ i in the direction from the + to - the plate of the capacitor

mg

The conduction current flows from the positive to the negative plate of the capacitor

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Thank you for your attentions!
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