Download 27.1 Magnetic Forces

Physics 4
Magnetic Forces and Fields
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What creates a magnetic field?
Answer: MOVING CHARGES
What is affected by a magnetic field?
Answer: MOVING CHARGES
We have a formula for magnetic force on a moving charge:

 
Fmag  q  v  B  q  v  B  sin()
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What creates a magnetic field?
Answer: MOVING CHARGES
What is affected by a magnetic field?
Answer: MOVING CHARGES
We have a formula for magnetic force on a moving charge:

 
Fmag  q  v  B  q  v  B  sin()
This is a vector cross-product. We need a right-hand-rule to find the direction of this force.
How to find the direction:
0)Use your RIGHT HAND
1)Fingers start in the direction of the charge’s velocity.
2)Curl fingers toward the direction of the magnetic field.
3)Thumb points in the direction of the magnetic force on the charge.
4)If the charge is negative, flip your hand over.
Important notes: If velocity is aligned with the magnetic field, force is zero.
Magnetic force is always perpendicular to both the velocityPrepared
and the
B-field.
by Vince
Zaccone
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Assistance Services at UCSB
A uniform magnetic field is directed into the page.
Find the direction of the magnetic force on charge A, a positive charge.
X
 X
B
X
X
X
X
X
X
X

vA
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
A
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A uniform magnetic field is directed into the page.
Find the direction of the magnetic force on charge A, a positive charge.
X
 X
B
X
X
X
X

vA
X
X
X
Fmag
A
-Fingers start along A’s velocity (up in the picture)
X
X
X
-Fingers bend toward B-field (into the page)
-Thumb sticks out to the left.
X
X
X
X
X
X
X
X
X
X
X
X
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A uniform magnetic field is directed into the page.
Find the direction of the magnetic force on charge B, a positive charge.
X
X
 X
B
X

vB
X
X
X
X
X
X
X
X
B
X
X
X
X
X
X
X
X
X
X
X
X
Prepared by Vince Zaccone
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A uniform magnetic field is directed into the page.
Find the direction of the magnetic force on charge B, a positive charge.
X
X
 X
B
X
X

vB
X
X
X
X
X
X
X
-Fingers start along B’s velocity (up and to the left
in the picture)
-Fingers bend toward B-field (into the page)
-Thumb sticks out down and to the left.
B
X
X

X
Fmag
X
X
X
X
X
X
X
X
X
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A uniform magnetic field is directed into the page.
Find the direction of the magnetic force on charge C, a negative charge.
X
 X
B
X
X
X
X
X
X
X
X
X
X

vC
C
X
X
X
X
X
X
X
X
X
X
X
X
Prepared by Vince Zaccone
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A uniform magnetic field is directed into the page.
Find the direction of the magnetic force on charge C, a negative charge.
X
X
 X
B
X
X
X
X
X
X
X
X
X

vC
C
-Fingers start along C’s velocity (to the right in the
picture)
-Fingers bend toward B-field (into the page)
-Thumb sticks out upward.
X
X
X
X

Fmag
X
X
X
X
X
X
X
X
-Force is downward. (negative charges are
pushed in the opposite direction)
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A uniform magnetic field is directed into the page.
Now that we can find the force on a charge, we should be able to predict its trajectory.
Notice that the force is always perpendicular to the velocity. This will yield a circular
path. In other words, the magnetic force is a centripetal force.
X
 X
B
X
X
X
X
Notice that in this diagram, A is a positive
charge, and moves counter-clockwise.

vA
X
X
X
Fmag
A
X
X
X
X
X
X
X

vC
C
X
X
Since B is a negative charge, it moves the
opposite direction - clockwise.
So you can tell the sign of the charge by which
direction it rotates in a magnetic field.
X

Fmag
X
X
X
X
What path would a neutral charge follow?
X
Prepared by Vince Zaccone
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Prepared by Vince Zaccone
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Application: Mass Spectrometer
Ions are fired into a region of constant magnetic field.
Magnetic force pushes it into a circular path and the impact location
will determine the mass/charge ratio of the ion.
𝐹𝑚𝑎𝑔 = 𝐹𝑐𝑒𝑛𝑡
𝑚𝑣 2
𝑞𝑣𝐵 =
𝑟
𝑚𝑣
𝑟=
𝑞𝐵
Velocity selector
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Application: Mass Spectrometer
Here is a problem from the book:
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Application: Mass Spectrometer
Here is a problem from the book:
We can use the radius formula we just
derived. How do we get the potential
difference involved?
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Application: Mass Spectrometer
Here is a problem from the book:
We can use the radius formula we just
derived. How do we get the potential
difference involved?
Recall the formula for electrical potential
energy: Uelec=qV
Since they are accelerated from rest, this
electrical potential energy becomes the kinetic
energy of the particle.
Prepared by Vince Zaccone
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Assistance Services at UCSB
Application: Mass Spectrometer
Here is a problem from the book:
We can use the radius formula we just
derived. How do we get the potential
difference involved?
Recall the formula for electrical potential
energy: Uelec=qV
Since they are accelerated from rest, this
electrical potential energy becomes the kinetic
energy of the particle.
The charge on each particle is just e, the
elementary charge. So we can write a formula:
1
2
2𝑚𝑣
= 𝑒𝑉
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Application: Mass Spectrometer
Here is a problem from the book:
We can use the radius formula we just
derived. How do we get the potential
difference involved?
Recall the formula for electrical potential
energy: Uelec=qV
Since they are accelerated from rest, this
electrical potential energy becomes the kinetic
energy of the particle.
The charge on each particle is just e, the
elementary charge. So we can write a formula:
1
2
2𝑚𝑣
= 𝑒𝑉
Rearrange to solve for velocity:
𝑣=
2𝑒𝑉
𝑚
And plug back into our radius formula:
𝑚
𝑟=
2𝑒𝑉
𝑚
𝑒𝐵
=
2𝑒𝑉𝑚
𝑒𝐵
Subtracting the radius for the two isotopes
gives their desired result.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Application: Mass Spectrometer
Here is a problem from the book:
We can use the radius formula we just
derived. How do we get the potential
difference involved?
Recall the formula for electrical potential
energy: Uelec=qV
Since they are accelerated from rest, this
electrical potential energy becomes the kinetic
energy of the particle.
The charge on each particle is just e, the
elementary charge. So we can write a formula:
1
2
2𝑚𝑣
= 𝑒𝑉
Rearrange to solve for velocity:
𝑣=
2𝑒𝑉
𝑚
And plug back into our radius formula:
𝑚
𝑟=
2𝑒𝑉
𝑚
𝑒𝐵
Part b) is just plugging in numbers.
Answer: V = 3317 Volts
=
2𝑒𝑉𝑚
𝑒𝐵
Subtracting the radius for the two isotopes
gives their desired result.
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A uniform magnetic field is directed into the page.
Find the direction of the magnetic force on the wire, with current flowing as shown.
X
 X
B
X
X
X
X
I
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
A uniform magnetic field is directed into the page.
Find the direction of the magnetic force on the wire, with current flowing as shown.
X
 X
B
X
X
X
X
I
X
X
X
Fmag
X
Right-Hand Rule (same as before)
X
X
Pretend that positive charges are flowing in the
direction of the current.
-Fingers start along A’s current (up in the picture)
-Fingers bend toward B-field (into the page)
X
X
X
X
X
X
X
X
X
X
X
X
-Thumb sticks out to the left.
Here is a formula for the force
on a wire in a B-field:
𝐹 = 𝐼 𝑙 × 𝐵 = 𝐼𝑙𝐵 sin 𝜃
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A uniform magnetic field is directed into the page.
Find the direction of the magnetic force on the wire, with current flowing as shown.
X
 X
B
X
X
X
X
I
X
X
X
Fmag
X
Right-Hand Rule (same as before)
X
X
Pretend that positive charges are flowing in the
direction of the current.
-Fingers start along A’s current (up in the picture)
-Fingers bend toward B-field (into the page)
X
X
X
X
X
X
X
X
X
X
X
X
-Thumb sticks out to the left.
Here is a formula for the force
on a wire in a B-field:
If your wire is not straight you can do
an integral along the wire instead:
𝐹 = 𝐼 𝑙 × 𝐵 = 𝐼𝑙𝐵 sin 𝜃
𝑑𝐹 = 𝐼𝑑 𝑙 × 𝐵 = 𝐼𝑙𝐵 sin 𝜃
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An example from the book. Try it (it’s a fun one – I promise!)
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An example from the book. Try it (it’s a fun one – I promise!)
The long way to do this problem involves integrating along the wire. This involves 3
sections – 2 straight parts and one semicircular arc.
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An example from the book. Try it (it’s a fun one – I promise!)
The long way to do this problem involves integrating along the wire. This involves 3
sections – 2 straight parts and one semicircular arc.
The shortcut uses symmetry to our advantage. The force on each section of wire is
perpendicular to the wire. If we break the force into components, the vertical
components cancel, leaving just the horizontal components.
It comes out the same as a wire that is straight instead of curved. This saves us the
trouble of integrating and we can just multiply it out.
𝐹 = 𝐼𝑙𝐵 sin 𝜃 = 3.40𝐴 3.00𝑚 2.20𝑇 sin 90° = 22.4𝑁
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Application: Torque on a current loop
This is the basis for electric motors.
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Application: Torque on a current loop
Here is a problem from the book:
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Application: Torque on a current loop
Here is a problem from the book:
The net force will be zero in both parts. Each side of
the loop feels a magnetic force, and they cancel.
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Application: Torque on a current loop
Here is a problem from the book:
The net force will be zero in both parts. Each side of
the loop feels a magnetic force, and they cancel.
As for the torque, in part a) we can see from the
diagram that the coil is in an unstable equilibrium
position. The field lines go straight through the loop,
but the direction of the current makes the dipole
moment of the coil point out of the page (opposite
the field). In the book’s explanation, the angle
φ=180°. So the torque is zero.
If the current in the wire went the opposite way, the
coil would be in a stable equilibrium.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Application: Torque on a current loop
Here is a problem from the book:
The net force will be zero in both parts. Each side of
the loop feels a magnetic force, and they cancel.
As for the torque, in part a) we can see from the
diagram that the coil is in an unstable equilibrium
position. The field lines go straight through the loop,
but the direction of the current makes the dipole
moment of the coil point out of the page (opposite
the field). In the book’s explanation, the angle
φ=180°. So the torque is zero.
If the current in the wire went the opposite way, the
coil would be in a stable equilibrium.
For part b) we will have to do some calculating.
Tilting the loop by 30° makes the angle φ=150°.
Using the book’s formula, we get:
𝜏 = 𝐼𝐴𝐵 sin 𝜑 = 1.40𝐴 0.22𝑚 0.35𝑚 1.50𝑇 sin 150° = 0.08 𝑁 ∙ 𝑚
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Application: Torque on a current loop
Here is a problem from the book:
The net force will be zero in both parts. Each side of
the loop feels a magnetic force, and they cancel.
As for the torque, in part a) we can see from the
diagram that the coil is in an unstable equilibrium
position. The field lines go straight through the loop,
but the direction of the current makes the dipole
moment of the coil point out of the page (opposite
the field). In the book’s explanation, the angle
φ=180°. So the torque is zero.
If the current in the wire went the opposite way, the
coil would be in a stable equilibrium.
For part b) we will have to do some calculating.
Tilting the loop by 30° makes the angle φ=150°.
Using the book’s formula, we get:
𝜏 = 𝐼𝐴𝐵 sin 𝜑 = 1.40𝐴 0.22𝑚 0.35𝑚 1.50𝑇 sin 150° = 0.08 𝑁 ∙ 𝑚
This is the magnitude. The direction of the torque is along the axis (up in their diagram).
The torque will always be oriented in such a way as to rotate the coil so that its dipole moment
aligns with the field. In this problem that means the coil needs to flip all the way over.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Application: Torque on a current loop
Here is a problem from the book:
The net force will be zero in both parts. Each side of
the loop feels a magnetic force, and they cancel.
As for the torque, in part a) we can see from the
diagram that the coil is in an unstable equilibrium
position. The field lines go straight through the loop,
but the direction of the current makes the dipole
moment of the coil point out of the page (opposite
the field). In the book’s explanation, the angle
φ=180°. So the torque is zero.
If the current in the wire went the opposite way, the
coil would be in a stable equilibrium.
For part b) we will have to do some calculating.
Tilting the loop by 30° makes the angle φ=150°.
Using the book’s formula, we get:
𝜏 = 𝐼𝐴𝐵 sin 𝜑 = 1.40𝐴 0.22𝑚 0.35𝑚 1.50𝑇 sin 150° = 0.08 𝑁 ∙ 𝑚
This is the magnitude. The direction of the torque is along the axis (up in their diagram).
The torque will always be oriented in such a way as to rotate the coil so that its dipole moment
aligns with the field. In this problem that means the coil needs to flip all the way over.
You can probably imagine what would happen if the loop was allowed to rotate – it
would flip over, overshoot, then flip back and forth, oscillating about the stable
equilibrium point. This is *almost* an electric motor. What needs to be done to
make it keep rotating is to reverse the current every time it flips so that the torque
always points the same way.
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