Physics 4
Magnetic Induction
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Magnetic Flux
Before we can talk about induction we need to understand magnetic flux.
You can think of flux as the number of field lines passing through an area.
Here is the formula:


flux   B   B  dA   B  dA  cos( )
If the field is uniform and the area is flat, we have:
 
flux   B  B  A  B  A  cos( )
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Faraday’s Law of Induction
 
flux   B  B  A  B  A  cos( )
Note that there are 3 parts to this formula: field strength, Area, and angle.
We will learn that magnetic induction occurs if any of these change.
In fact, the rate of change of flux is the induced voltage:
Here is Faraday’s Law of Induction:
 induced  
d B
dt
The negative sign helps to remind us of what induction does –
induction always opposes the change in the flux. We call this Lenz’s Law.
Next we will look at a few specific examples of a changing magnetic flux:
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1) A loop of wire entering a uniform magnetic field.
X
X
X
X

B
X
X
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X
v
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X
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X
Question: which direction is the induced current in the loop?
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1) A loop of wire entering a uniform magnetic field.
X
X
X
X

B
X
X
X
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X
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X
I induced
B induced
v
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X
As the loop enters the B-field, the flux increases into the page, so there will be a current induced in the loop to
oppose this increasing flux. The direction of the induced current is Counter-Clockwise, because that direction will
create an induced B-field through the loop, out of the page (opposing the increase into the page).
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1) A loop of wire entering a uniform magnetic field.
X
X
X
X

B
X
X
X
X
X
X
X
X
I induced
B induced
v
X
X
X
X
X
X
X
X
X
X
X
X
You might also notice that there will be a net force on the loop to the right, opposing its motion (try the right hand
rule on each side of the loop). Induction opposes the change by trying to push the loop back out of the B-field.
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2) A loop of wire rotating in a uniform magnetic field.
X
X
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X
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X
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X
X

B
Question: which direction is the induced current in the loop?
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2) A loop of wire rotating in a uniform magnetic field.
X
X
X
X

B
X
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X B induced
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I induced
X
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X
As shown, the loop is intercepting the maximum flux. As it rotates, the flux will be decreasing, so there will be an
induced current to oppose this change. The direction of the current will be clockwise, which creates an induced
magnetic field through the loop, into the page.
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3) A stationary loop of wire in a magnetic field that is increasing in strength.
X
B(increa sin g)
X
X
X
X
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X
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X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
Question: which direction is the induced current in the loop?
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3) A stationary loop of wire in a magnetic field that is increasing in strength.
X
X
B(increa sin g)
X
X
X
X
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X
I induced
B induced
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X
The increasing B-field causes the flux to increase, inducing current in the loop. The direction of the current is
Counter-Clockwise, creating an induced B-field out of the page through the loop, opposing the change in the flux.
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4) A bar magnet approaching a stationary loop of wire.
N
S
Question: which direction is the induced current in the loop, as viewed from the right?
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4) A bar magnet approaching a stationary loop of wire.
Iinduced
N
S

Bmagnet

Binduced
View from the right
X
X
X

Bmagnet
X
X

Binduced
X
I induced
When viewed from the right, this looks like a circular loop with an increasing (because the magnet is approaching)
B-field directed through the loop, to the left. The induced current will oppose this increasing field by creating some
flux in the opposite direction (to the right, or out of the page in the view from the right). This means the induced
current must be Counter-Clockwise.
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5) A rotating loop of wire near a stationary bar magnet.
N
S
Question: which direction is the induced current in the loop, as viewed from the right?
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5) A rotating loop of wire near a stationary bar magnet.
View from the right
N
S

Bmagnet

Binduced
X
X
Iinduced
X

Bmagnet
X
X X
Binduced
X
I induced
As shown, the loop is intercepting maximum flux, so when it rotates, the flux decreases. Induction will create
some flux through the loop to the right, to oppose the change. This means the induced current is Clockwise.
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6) A loop of wire approaching a straight wire with a steady current.
Current = I (steady)
Question: which direction is the induced current in the loop?
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6) A loop of wire approaching a straight wire with a steady current.
I induced
X
B wire
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X
Binduced
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Current = I (steady)
In the vicinity of the loop, the wire creates a magnetic field directed into the page. As the loop approaches, the
field gets stronger, so the flux is increasing. To oppose this, the induced current must be Counter-Clockwise to
create some induced flux out of the page.
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7) A stationary loop of wire near a straight wire with decreasing current.
Current = I (decreasing)
Question: which direction is the induced current in the loop?
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7) A stationary loop of wire near a straight wire with decreasing current.
I induced
X
B wire
X
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X
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X
X
X
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X
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
B
X induced
Current = I (decreasing)
In the vicinity of the loop, the wire creates a magnetic field directed into the page. As the current decreases, the
field gets weaker, so the flux is decreasing. To oppose this, the induced current must be Clockwise to create
some induced flux into the page.
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Here’s one where we can calculate the effect of the induction:
Suppose there is a uniform B-Field pointing into the page. A circuit is placed in the field so that it intercepts the
maximum magnetic flux. One side of the rectangular circuit is free to slide back and forth along the rails. Let’s
first try to figure out what happens qualitatively:
R
V
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Here’s one where we can calculate the effect of the induction:
Suppose there is a uniform B-Field pointing into the page. A circuit is placed in the field so that it intercepts the
maximum magnetic flux. One side of the rectangular circuit is free to slide back and forth along the rails. Let’s
first try to figure out what happens qualitatively:
Iinitial
R
V
First we can find the direction of the current in the given loop. Current will flow clockwise. In the diagram it is
labeled Iinitial.
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Here’s one where we can calculate the effect of the induction:
Suppose there is a uniform B-Field pointing into the page. A circuit is placed in the field so that it intercepts the
maximum magnetic flux. One side of the rectangular circuit is free to slide back and forth along the rails. Let’s
first try to figure out what happens qualitatively:
Iinitial
R
Fmag
V
First we can find the direction of the current in the given loop. Current will flow clockwise. In the diagram it is
labeled Iinitial.
The force on the sliding wire can also be found from the right hand rule for magnetic forces. This force points to
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the right. So the wire will slide to the right.
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Here’s one where we can calculate the effect of the induction:
Suppose there is a uniform B-Field pointing into the page. A circuit is placed in the field so that it intercepts the
maximum magnetic flux. One side of the rectangular circuit is free to slide back and forth along the rails. Let’s
first try to figure out what happens qualitatively:
Iinitial
R
Fmag
V
Now that the wire is moving, we can talk about the induction effect (remember, induction only happens when the
flux is changing). The magnetic flux through the loop is increasing into the page, and there will be an induced
current in the wire that opposes this change.
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Here’s one where we can calculate the effect of the induction:
Suppose there is a uniform B-Field pointing into the page. A circuit is placed in the field so that it intercepts the
maximum magnetic flux. One side of the rectangular circuit is free to slide back and forth along the rails. Let’s
first try to figure out what happens qualitatively:
Iinitial
R
V
Fmag
Iinduced
Now that the wire is moving, we can talk about the induction effect (remember, induction only happens when the
flux is changing). The magnetic flux through the loop is increasing into the page, and there will be an induced
current in the wire that opposes this change.
This induced current will flow counter-clockwise, creating some induced flux to oppose the increasing flux.
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Here’s one where we can calculate the effect of the induction:
Suppose there is a uniform B-Field pointing into the page. A circuit is placed in the field so that it intercepts the
maximum magnetic flux. One side of the rectangular circuit is free to slide back and forth along the rails. Let’s
first try to figure out what happens qualitatively:
Iinitial
R
V
Finduced
Fmag
Iinduced
Now that the wire is moving, we can talk about the induction effect (remember, induction only happens when the
flux is changing). The magnetic flux through the loop is increasing into the page, and there will be an induced
current in the wire that opposes this change.
This induced current will flow counter-clockwise, creating some induced flux to oppose the increasing flux.
Notice – this also creates an induced force on the wire to the left.
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Here’s one where we can calculate the effect of the induction:
Suppose there is a uniform B-Field pointing into the page. A circuit is placed in the field so that it intercepts the
maximum magnetic flux. One side of the rectangular circuit is free to slide back and forth along the rails. Let’s
first try to figure out what happens qualitatively:
Iinitial
R
V
Finduced
Fmag
Iinduced
So the wire initially accelerates to the right, but the faster it moves, the stronger the induction, and the closer it
gets to some maximum speed (where the induced force will match up with the initial Fmag. We can try to calculate
this maximum speed.
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x
L
Iinitial
R
V
Finduced
Fmag
Iinduced
To get the max speed, we will have to use our formula for the induced voltage. As soon as the induced voltage
matches up with the battery voltage we will be at equilibrium, and the wire will be moving at maximum speed.
Vinduced 
B (B  A  cos )

t
t
 x 
Vinduced  B  L     B  L  v
 t 
B  L  v max  Vbattery
v max 
Vbattery
B L
In this formula, the only thing changing is the horizontal distance
which is labeled x in the diagram above. If we replace the Area
with L*x and realize that we can ignore the angle (90 degrees), we
get a simpler formula, involving the velocity of the wire.
Setting this induced voltage equal to the battery voltage, we will get
the maximum speed. Another option would be to set Finduced = Fmag.
Should get the same result, after using Ohm’s Law. Prepared by Vince Zaccone
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Example: In the figure the loop is being pulled the right at constant
speed v, and there is a constant current I in the straight wire.
Find the emf induced in the loop, and the direction of the induced current.
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Example: In the figure the loop is being pulled the right at constant
speed v, and there is a constant current I in the straight wire.
Find the emf induced in the loop, and the direction of the induced current.
We will use Faraday’s law of induction for this one. The flux
through the loop is due to the field produced by the wire, which
has a formula:
𝜇0 𝐼
𝐵𝑤𝑖𝑟𝑒 =
2𝜋𝑟
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Example: In the figure the loop is being pulled the right at constant
speed v, and there is a constant current I in the straight wire.
Find the emf induced in the loop, and the direction of the induced current.
We will use Faraday’s law of induction for this one. The flux
through the loop is due to the field produced by the wire, which
has a formula:
𝜇0 𝐼
𝐵𝑤𝑖𝑟𝑒 =
2𝜋𝑟
This field is not constant, so we will have to integrate to get the
total flux through the loop:
Φ𝐵 =
𝐵 ∙ 𝑑𝐴
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Example: In the figure the loop is being pulled the right at constant
speed v, and there is a constant current I in the straight wire.
Find the emf induced in the loop, and the direction of the induced current.
We will use Faraday’s law of induction for this one. The flux
through the loop is due to the field produced by the wire, which
has a formula:
𝜇0 𝐼
𝐵𝑤𝑖𝑟𝑒 =
2𝜋𝑟
This field is not constant, so we will have to integrate to get the
total flux through the loop:
Φ𝐵 =
The B-field points into the page, which is straight
through the loop, so the dot product is just the
magnitudes multiplied together.
𝐵 ∙ 𝑑𝐴
𝑟+𝑎
Φ𝐵 =
𝑟
𝜇0 𝐼
𝜇0 𝐼𝑏
𝑏𝑑𝑟 =
2𝜋𝑟
2𝜋
𝑟+𝑎
𝑟
Limits on our integral are from r to r+a
Area of this
slice is dA=b·dr
𝑑𝑟
𝜇0 𝐼𝑏
𝑟+𝑎
𝑑𝑟 =
𝑙𝑛
𝑟
2𝜋
𝑟
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Example: In the figure the loop is being pulled the right at constant
speed v, and there is a constant current I in the straight wire.
Find the emf induced in the loop, and the direction of the induced current.
We will use Faraday’s law of induction for this one. The flux
through the loop is due to the field produced by the wire, which
has a formula:
𝜇0 𝐼
𝐵𝑤𝑖𝑟𝑒 =
2𝜋𝑟
This field is not constant, so we will have to integrate to get the
total flux through the loop:
Φ𝐵 =
The B-field points into the page, which is straight
through the loop, so the dot product is just the
magnitudes multiplied together.
𝐵 ∙ 𝑑𝐴
𝑟+𝑎
Φ𝐵 =
𝑟
𝜇0 𝐼
𝜇0 𝐼𝑏
𝑏𝑑𝑟 =
2𝜋𝑟
2𝜋
𝑟+𝑎
𝑟
Area of this
slice is dA=b·dr
Limits on our integral are from r to r+a
𝑑𝑟
𝜇0 𝐼𝑏
𝑟+𝑎
𝑑𝑟 =
𝑙𝑛
𝑟
2𝜋
𝑟
For Faraday’s law we take the derivative to find the induced emf:
𝜀𝑖𝑛𝑑𝑢𝑐𝑒𝑑 = −
𝑑Φ𝐵
𝑑𝑡
The distance r is changing at speed v as the loop is pulled away, so dr/dt = v
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Example: In the figure the loop is being pulled the right at constant
speed v, and there is a constant current I in the straight wire.
Find the emf induced in the loop, and the direction of the induced current.
We will use Faraday’s law of induction for this one. The flux
through the loop is due to the field produced by the wire, which
has a formula:
𝜇0 𝐼
𝐵𝑤𝑖𝑟𝑒 =
2𝜋𝑟
This field is not constant, so we will have to integrate to get the
total flux through the loop:
Φ𝐵 =
The B-field points into the page, which is straight
through the loop, so the dot product is just the
magnitudes multiplied together.
𝐵 ∙ 𝑑𝐴
𝑟+𝑎
Φ𝐵 =
𝑟
𝜇0 𝐼
𝜇0 𝐼𝑏
𝑏𝑑𝑟 =
2𝜋𝑟
2𝜋
𝑟+𝑎
𝑟
Area of this
slice is dA=b·dr
Limits on our integral are from r to r+a
𝑑𝑟
𝜇0 𝐼𝑏
𝑟+𝑎
𝑑𝑟 =
𝑙𝑛
𝑟
2𝜋
𝑟
For Faraday’s law we take the derivative to find the induced emf:
𝜀𝑖𝑛𝑑𝑢𝑐𝑒𝑑 = −
𝜀𝑖𝑛𝑑𝑢𝑐𝑒𝑑
𝑑Φ𝐵
𝑑𝑡
𝜇0 𝐼𝑎𝑏𝑣
=
2𝜋𝑟(𝑟 + 𝑎)
The distance r is changing at speed v as the loop is pulled away, so dr/dt = v
The current direction will oppose the change in the flux.
Since the flux decreases as the loop is pulled away, the
induced current is clockwise – creating an induced B-field
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into the page, as desired.
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Displacement Current
In a region where there is a changing electric
flux, we find that there is a magnetic field
produced. The field generated is the same as if
a current flowed through the region, even
though no wires are present there. We have a
formula for this displacement current:
𝑑Φ𝐸
𝑖𝑑 = 𝜖
𝑑𝑡
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Maxwell’s Equations
Maxwell’s equations summarize the relationships
between electric and magnetic fields:
𝐸 ∙ 𝑑𝐴 =
𝑄𝑒𝑛𝑐𝑙
𝜖0
𝐵 ∙ 𝑑𝐴 = 0
𝑮𝒂𝒖𝒔𝒔′ 𝒔 𝑳𝒂𝒘 𝒇𝒐𝒓 𝑬
𝑮𝒂𝒖𝒔𝒔′ 𝒔 𝑳𝒂𝒘 𝒇𝒐𝒓 𝑩
𝐵 ∙ 𝑑𝑙 = 𝜇0 𝑖𝑐 + 𝜖0
𝑑Φ𝐵
𝐸 ∙ 𝑑𝑙 = −
𝑑𝑡
𝑑Φ𝐸
𝑑𝑡
𝑨𝒎𝒑𝒆𝒓𝒆′ 𝒔 𝑳𝒂𝒘
𝑒𝑛𝑐𝑙
𝑭𝒂𝒓𝒂𝒅𝒂𝒚′𝒔 𝑳𝒂𝒘
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