1. Chapter 29 Magnetic Fields
Magnetic Fields and Forces
Magnetic Force Acting on a Current-Carrying Conductor
Motion of a Charged Particle in a Uniform Magnetic Field
Applications Involving Charged Particles Moving in aMagnetic Field
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2. Magnetism Magnetic Attraction-Repulsion
Since ancient times, certain materials, called magnets, have been known to have the property of attracting tiny pieces of metal. This attractive property is called magnetism.
The strength of a magnet is concentrated at the ends, called north and south “poles” of the magnet.
A suspended magnet: N-seeking end and S-seeking end are N and S poles.
Magnetic Attraction-Repulsion
Magnetic Forces: Like Poles Repel
Unlike Poles Attract
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3. Magnetic Field Lines
We can describe magnetic field lines by imagining a tiny compass placed at nearby points.
The direction of the magnetic field B at any point is the same as the direction indicated by this compass.
Field B is strong where lines are dense and weak where lines are sparse.
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4. Properties of Magnetic Field Lines
Magnetic lines of force never intersect.
By convention, magnetic lines of force point from north to south outside a magnet (and from south to north inside a magnet).
Field lines converge where the magnetic force is strong, and spread out where it is weak. (Number of lines per unit area is proportional to the magnetic field strength.)
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5. Force Due to Magnetic Field
The force exerted on a charged particle by a magnetic field is given by the vector cross product:
F = q v  B
F = force (vector)
q = charge on the particle (scalar)
v = velocity of the particle relative to field (vector)
B = magnetic field (vector) V q B
Recall that the magnitude of a cross is the product of the magnitudes of the vectors times the sine of the angle between them. So, the magnitude of the magnetic force is given by
F = q v B sin
where  is angle between q v and B vectors.
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6. Force on a moving charge in a magnetic field
The force on a moving charge is proportional to the component of the magnetic field perpendicular to the direction of the velocity of the charge and is in a direction perpendicular to both the velocity and the field.
The force is greatest when the velocity v is perpendicular to the B field. The deflection decreases to zero for parallel motion.
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7. Force is perpendicular to B,v
The magnetic force on a charge depends on the magnitude of the charge, its velocity, and the magnetic field F = q v B sin(θ)
Direction from RHR
Note if v is parallel to B then F=0 V q B
. If a charge moves parallel to a magnetic field, there is no magnetic force on it, since sin 0° = 0.
F = q v B sin
1 N = 1 C (m / s) (T)
From the formula for magnetic force we can find a relationship between the tesla and other SI units. The sin of an angle has no units, so
1 N
1 N =
1 T =
C (m / s)
A m
A magnetic field of one tesla is very powerful magnetic field. Sometimes it may be convenient to use the gauss, which is equal to tesla.
Force is perpendicular to B,v
B does no work! (W=F d cos q )
Speed is constant (W=D K.E. )
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8. Magnetic Field B:
A magnetic field intensity of one tesla (T) exists in a region of space where a charge of one coulomb (C) moving at 1 m/s perpendicular to the B-field will experience a force of one newton (N).
Each of the following results in a greater magnetic force F: an increase in velocity v, an increase in charge q, and a larger magnetic field B.
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9. Magnetic forces on moving charges
When a charge moves through a B-field, it feels a force.
It has to be moving
It has to be moving across the B-field
Direction of force is determined by another right-hand rule
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10. Right Hand Rule
Direction of force on a positive charge given by the right hand rule.
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11. = (y1 z2 - y2 z1) i - (x1 z2 - x2 z1) j + (x1 y2 - x2 y1) k
Let v1 =  x1, y1, z1 
and v2 =  x2, y2, z2 .
v1  v2 =
x1 y1 z1
x2 y2 z2
i j k
= (y1 z2 - y2 z1) i - (x1 z2 - x2 z1) j + (x1 y2 - x2 y1) k
Note that the cross product of two vectors is a vector itself that is  to each of the original vectors. i, j, and k are the unit vectors pointing, along the positive x, y, and z axes, respectively. (See the vector presentation for a review of determinants.)
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12. An electron moves perpendicular to a magnetic field
An electron moves perpendicular to a magnetic field. What is the direction of ?
1. Left
2. Up
3. Down
4. Into the page
5. Out of the page
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13. F = q v  B  F = q v B since sin 90° = 1.
This magnet is similar to a parallel plate capacitor in that there is a strong uniform field between its poles with some fringing on the sides. Suppose the magnetic field strength inside is 0.07 T and a 4.3 mC charge is moving through the field at right angle to the field lines. How strong and which way is the magnetic force on the
F = q v  B  F = q v B since sin 90° = 1.
So, F = N directed out of the page.
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14. Example
A proton is moving at a speed of 3 x 104 m/s towards the West through a magnetic field of strength 500 Gauss directed South. What is the strength and direction of the magnetic force on the proton at thi instant?
qproton = +e = 1.6 x C.
v = 3 x 104 m/s, West
B = 500 Gauss * 1 Tesla/10,000 Gauss = .05 T, South
magnitude: Fmagnetic = q v B sin(θ)
direction: right hand rule
magnitude: F = (1.6 x C) (3 x 104 m/s) * (.05 T) * sin(90o) = 2.4 x Nt.
direction: thumb = hand x fingers
= West x South = UP.
Note: although the force looks small, consider the acceleration: a = F/m =
2.4 x Nt / 1.67 x kg = 1.44 x 1011 m/s2.
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15. Example
A 2-nC charge is projected with velocity 5 x 104 m/s at an angle of 300 with a 3 mT magnetic field as shown. What are the magnitude and direction of the resulting force?
Draw a rough sketch.
v sin f v
300 B v F B
q = 2 x 10-9 C v = 5 x 104 m/s B = 3 x 10-3 T q = 300
Fmagnetic = q v B sin(θ)
Using right-hand rule, the force is seen to be upward.
Resultant Magnetic Force: F = 1.50 x 10-7 N, upward
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16. The magnetic force magnitude:
A uniform B field 1.2 mT points vertically upwards. A 5.3 MeV proton moves horizontally from south to north. What is the magnetic force on the proton
[proton mass mp = 1.6 x kg, 1 eV = 1.6 x J]
Firstly, find velocity of the proton:
K = (1/2) mv2 = 5.3 MeV = (5.3 x 106) x (1.6 x 10-19)  J
the velocity: 
The magnetic force magnitude:
Direction is given by the right hand rule: to the East
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19. What is the direction of the force F on the charge in each of the examples described below?+ v Up F
X X X X X X X X X X X X X X X X
X X X X X X X X X X X X X X X X
Left F v +
· · · · Up F
· · · ·
negative q - v F
Right - v
negative q
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20. Direction of Magnetic Force on Moving Charges
Velocity B Force
out of page right up
out of page left down
out of page up
1) Up 2) Down 3) Right 4) Left ) Zero
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21. Velocity B Force out of page right up out of page left down
out of page up
out of page down
left
right
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23. An electron moves in the magnetic field B=0. 50 i T with a speed of 1
An electron moves in the magnetic field B=0.50 i T with a speed of 1.0x107m/s in the direction shown. What is the magnetic force (in component form) on the electron?
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24. Direction RHR
Direction RHR
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25. A) negative z direction B) positive y direction
The figure shows a charged particle with velocity v travels through a uniform magnetic field B. What is the direction of the magnetic force F on the particle?
A) negative z direction
B) positive y direction
C) positive z direction
D) negative x direction
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27. A) positive y direction B) positive z direction C) the force is 0
The figure shows a charged particle with velocity v travels through a uniform magnetic field B. What is the direction of the magnetic force F on the particle?
A) positive y direction
B) positive z direction
C) the force is 0
D) negative x direction
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28. A) positive y direction B) the force is 0 C) negative z direction
The figure shows a charged particle with velocity v travels through a uniform magnetic field B. What is the direction of the magnetic force F on the particle?
A) positive y direction
B) the force is 0
C) negative z direction
D) negative x direction
F = q v B sin(180) = 0
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29. A section of wire carrying current to the right is shown in a uniform magnetic field. We can imagine positive charges moving to right, each feeling a magnetic force out of the page. This will cause the wire to bow outwards. Shown on the right is the view as seen when looking at the N pole from above. The dots represent a uniform
mag. field coming out of the page. The mag. force on the wire is proportional to the field strength, the current, and the length of the wire. B
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30. Particle is moving straight upwards then veers to the right.
Each chamber has a unique magnetic field. A positively charged particle enters chamber 1 with velocity 75 m/s up, and follows the dashed trajectory. 1 2
v = 75 m/s
q = +25 mC
What is the direction of the force on the particle just as it enters region 1?
1) up
2) down
3) left
4) right
5) into page
6) out of page
Particle is moving straight upwards then veers to the right.
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31. What is the direction of the magnetic field in region 1? 1) up 2) down
Each chamber has a unique magnetic field. A positively charged particle enters chamber 1 with velocity 75 m/s up, and follows the dashed trajectory. 1 2
v = 75 m/s
q = +25 mC
What is the direction of the magnetic field in region 1?
1) up
2) down
3) left
4) right
5) into page
6) out of page
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32. Motion of a Charge in a Magnetic Field
The ’s represent field lines pointing into the page. A positively charged particle of mass m and charge q is shot to the right with speed v. By the right hand rule the magnetic force on it is up. Since v is  to B, F = FB = q v B. Because F is  to v, it has no tangential component; it is entirely centripetal. Thus F causes a centripetal acceleration. As the particle turns so do v and F, and if B is uniform the particle moves in a circle. This is the basic idea behind a particle accelerator like Fermilab. Since F is a centripetal force, F = FC = m v2 / R. Let’s see how speed, mass, charge,
field strength, and radius of curvature are related:
FB = FC
 q v B = m v2 / R
m v
 R =
q B
Uniform B into page
Since v is at right angles to B, θ = 90° and sinθ = 1.
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33. Heavier masses will give bigger radii, but we can shrink the radii if they become too big by using bigger magnetic fields.
If the motion is exactly at right angles to a uniform field, the path is turned into a circle.
In general, with the motion inclined to the field, the path is helix round the lines of force.
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35. Circulating Charged Particles
The key feature of the magnetic force on a charged particle is that it is perpendicular to the particle's velocity.
Assume that a uniform field can be produced in the laboratory and that we can shoot charged particles through it. What will the trajectory be of such a particle?
Consider the following diagram. We let a positive charged particle have a velocity along the x-axis, and apply a uniform magnetic field along the y-axis:
The magnetic force will be along the +z axis, as shown below:
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36. First note that the magnetic force does no work on the particle
First note that the magnetic force does no work on the particle. The rate at which work is performed is given by the power:
But v and F are perpendicular - F is the cross product of v and B, and is thus perpendicular to both of these. The dot product of two perpendicular vectors is zero. Zero power means that no work is performed by this force. Since no work is performed, the kinetic energy and speed of the particle must be constant. You should recall that this is very different from the result of applying an electric field to a charged particle.
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37. Circulating Charged Particles
We therefore find that the force 1) is perpendicular to the velocity, and 2) does not change the kinetic energy. What kind of motion will result? Even though the speed is constant, the direction of the velocity certainly changes - the force and therefore the acceleration (F = ma) are not zero.
When the magnetic field is perpendicular to the velocity, the motion will clearly be circular since the force will always be oriented 'inward' perpendicular to the velocity - it is a centripetal force which leads to a centripetal acceleration. The trajectory, for a positive charge, is shown below:
It is a circle, when viewed from 'above' (from the +y axis) and the motion will be clockwise. If the particle were negatively charged, then the force would be in the opposite direction. That is, when the velocity is along the +x-axis and the field along the +y-axis, the force is along the -z-axis. In this case, the motion will be counter-clockwise when viewed from above:
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38. What happens if the velocity is not perpendicular to the field
What happens if the velocity is not perpendicular to the field? In this case, we can resolve the velocity into two components, one parallel to the field and one perpendicular. The magnetic force is associate only with the perpendicular component, and since cross products are distributive, we write the cross-products as:
The force will be perpendicular to the field and to the component of the velocity that is perpendicular to the field. This perpendicular component of velocity varies in the same way as the previous example of circular motion. There will be no force along the magnetic field, so the motion in that direction is simply a constant velocity given by the parallel component of velocity to the field - this component is unaffected by the magnetic field. In this case, the motion will be helical: B
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39. From this we can evaluate the radius of the orbit:
We know the magnitude of the magnetic force is qvB, where v is the component of velocity perpendicular to the field. So for simplicity, let us assume there is no velocity parallel to the field, so this velocity is really the particle's speed. Because the magnetic force is always directed toward the center of the orbit, we can set the acceleration equal to the centripetal acceleration for circular motion, v^2/r. We thus have that
From this we can evaluate the radius of the orbit:
The period T (the time for one full revolution) is equal to the circumference divided by the speed
The frequency (the number of revolutions per unit time) is
It should be noted that, the quantities T and f do not depend on the speed of the particle, however faster particles move in larger circles. All particles with the same q/m ratio take the same time T to complete one trip.
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40. Magnetic Force on a current carrying wire
We have already talked about how a magnetic field exerts a sidewise force on moving electrons in a conductor (Hall effect). This force is transmitted to the conductor itself, because the conduction electrons can not escape sideways out of the conductor.
These figures are from your book which clearly illustrates the direction of forces exerted on a wire with current in the presence of a B-field . In this case the B-field is directed out of the screen and the conducting wire is flexible. To the right in Figure (a) the wire has no current so no forces are present. The figure below shows what happens in the wire when current is sent upward through the wire. As the electrons move downward with drift velocity vd they are deflected to right with a magnetic force FB =e vd B on each electron. So we expect the wire to deflect to the right as shown in right Figure labeled (b)
Consider a length L of wire in the figure on the left. The conduction electrons in this section will drift past the plane x---x in the figure in a time t = L / vd Therefore in that time, a charge
will pass through this plane
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41. Magnetic Force on a current carrying wire
q = i L/vd
Recalling the magnetic force equation
FB = qvB sin
Substitute the drift velocity vd for v, the above equation for q and  = 90o , we have
This relationship gives the force that acts on a segment of a straight wire of length L, carrying a current i and immersed in magnetic field B that is perpendicular to the wire.
If the magnetic field is not perpendicular to the wire, we must consider general form of the above equation which is
L is a length vector that has a magnitude L and is in the direction of the current i.
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42. Magnetic Force on a current carrying wire
The magnitude of the force FB = i L x B can be written as
FB = i L B sin 
where  is the angle between the directions of L and B
Since the current is taken to be a positive quantity, the direction of FB is
that of the cross product L x B and FB is always perpendicular to the plane defined by the vectors L and B.
Recall the force equation that defined the B-field in terms of the force acting a single moving charge.
FB = q v x B
This equation is almost equivalent to the force on a wire equation
FB = i L x B
and in practice this equation generally preferred as the equation defining the B-field since its far easier doing measurements with currents than single charges
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43. Magnetic Force on a current carrying wire
If a wire is not straight, we can then consider breaking up the wire into small straight segments and apply the force equation
FB = I L x B
to each segment. The net force on the wire is the vector sum of the forces on each of the segments.
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44. Force on a Current-Carrying Wire
In a wire carrying current the charge carriers are moving at the drift velocity vd (on average),
what is the force on the wire?
F = ILB
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45. Force on a Current-Carrying Wire
In a wire carrying current the charge carriers are moving at the drift velocity vd (on average), and so in a magnetic field they each feel force F = qvdB
In Ch 27 we computed vd = I/Aqn where I = current, n = charge carriers per unit volume.
In the whole wire, the number of carrier is nAL, so for the situation shown the total force on the wire is
F = (nAL)q(I/Aqn)B = ILB
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46. Right hand rule gives direction
What is the direction?
Vector form: L
F = I LB
where L is a vector along the wire segment in the direction of positive current
Right hand rule gives direction
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47. Force: Ex. 2
A loop of wire carries A and is placed in a magnetic field. The loop is 10.0 cm wide and experiences a force of 3.48 X 10-2 N downward (on top of gravity). What is the strength of the magnetic field?
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48. F = IlBsinq F = IlBsin90o F = IlB(1) F = IlB B = F Il B = 3
F = IlBsinq F = IlBsin90o F = IlB(1) F = IlB B = F Il B = 3.48 X 10-2 N = 1.42 T (0.245 A)(0.100m)
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49. Example
To see if this is really feasible, let’s try using realistic numbers to see what the radius for a proton should be:
q = 1.6 x Coul; B = .05 Teslas (500 G)
m = 1.67 x kg; Vacc = 500 volts gives:
(1/2)mv2 = qV, or v = [2qV/m]1/2 = 3.1 x 105 m/s
r = mv/qB = (1.67 x 10-27kg)*(3.1 x 105 m/s) / (1.6 x Coul)*(0.05 T) = .065 m = 6.5 cm.
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50. A positive charge moves along a circular path under the influence of a magnetic field. The magnetic field is perpendicular to the plane of the circle, as in the Figure. If the velocity of the particle is reversed at some point along the path, the particle will not retrace its path. If the velocity of the particle is suddenly reversed, then from RHR-1 we see that the force on the particle reverses direction. The particle will travel on a different circle that intersects the point where the direction of the velocity changes. The direction of motion of the particle (clockwise or counterclockwise) will be the same as that in the original circle. This is suggested in the figure below.
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51. Example
A Neon ion, q = 1.6 x C, follows a path of radius 7.28 cm. Upper and lower B = 0.5 T and E = 1000 V/m. What is its mass? +q R + -
x x x x x x x x
Photographic plate m
slit
x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x
v = 2000 m/s
m v
R =
q B
m = 2.91 x kg
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52. Each chamber has a unique magnetic field
Each chamber has a unique magnetic field. A positively charged particle enters chamber 1 with velocity v1= 75 m/s up, and follows the dashed trajectory. 1 2
v = 75 m/s
q = +25 mC
What is the speed of the particle when it leaves chamber 2?
1) v2 < v1
2) v2 = v1
3) v2 > v1
Magnetic force is always perpendicular to velocity, so it changes direction, not speed of particle.
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53. Compare the magnitude of the magnetic field in chambers 1 and 2
Each chamber has a unique magnetic field. A positively charged particle enters chamber 1 with velocity v1= 75 m/s up, and follows the dashed trajectory. 1 2
v = 75 m/s
q = +25 mC
Compare the magnitude of the magnetic field in chambers 1 and 2
1) B1 > B2
2) B1 = B2.
3) B1 < B2
Larger B, greater force, smaller R
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54. Each chamber has a unique magnetic field
Each chamber has a unique magnetic field. A positively charged particle enters chamber 1 with velocity v1= 75 m/s up, and follows the dashed trajectory. 1 2
v = 75 m/s
q = ?? mC
A second particle with mass 2m enters the chamber and follows the same path as the particle with mass m and charge q=25 mC. What is its charge?
1) Q = 12.5 mC
2) Q = 25 mC
3) Q = 50 mC
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55. Either one sin(q) = sin(180-q)B
Which angle do you use to determine the magnitude of the force on the charged particle? (q1, q2 , either one).
Below I have drawn the trajectory for two charged particles traveling through a magnetic field. Is particle 1 positive or negatively charged?
If particles 1 and 2 have the same mass and velocity, which has the largest charge? V q1 q2
Either one sin(q) = sin(180-q) 1
Positive
Particle 1 2
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56. (a) out of the page (b) up (c) down (d) to right (e) to left
The charge carriers in the wire are electrons. What is the direction of the magnetic force on the wire?
(a) out of the page (b) up
(c) down (d) to right
(e) to left
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57. Magnetic interactions:
A moving charge or a current creates a magnetic field in the surrounding space
The magnitude field exerts a force on any other moving charge or current that is present in the field
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58. Example
A magnetron in a microwave oven emits electromagnetic waves with frequency f=2450MHz. What magnetic field strength is required for electrons to move in circular paths with this frequency?
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59. Velocity Selector Example
Determine magnitude and direction of magnetic field such that a positively charged particle with initial velocity v travels straight through and exits the other side. FB E v FE
Electric force is down, so need magnetic force up.
By RHR, B must be into page
For straight line, need |FE |= |FB |
q E= q v B sin(90)
B = E/v
What direction should B point if you want to select negative charges?
A) Into Page B) Out of page C) Left D) Right 38
FE would be up so FB must be down.
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60. E = vB Example. - E = (1.5 x 106 m/s)(20 x 10-3 T);
A lithium ion, q = +1.6 x C, is projected through a velocity selector where B = 20 mT. The E-field is adjusted to select a velocity of 1.5 x 106 m/s. What is the electric field E?
x x x x x x x x + - +q v
Source of +q V
E = vB
E = 3.00 x 104 V/m
E = (1.5 x 106 m/s)(20 x 10-3 T);
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61. Let’s find a relationship between q, B, and E if there is no deflection at all:
Need force up if proton is to be undeflected
Put magnetic field into page
Fnet = 0
 FB = FE E
 q v B = q E
 v = B
We want ay = 0
Then
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62. Example: Cathode Ray Tube
Consider the cathode-ray tube used for lecture demonstrations.
In this tube electrons form an electron beam when accelerated horizontally by a voltage of 136 V in an electron gun.
The mass of an electron is 10-31 kg while the elementary charge is 10-19 C.
(a) Calculate the velocity of the electrons in the beam after leaving the electron gun.
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63. Cathode Ray Tube (2)
(b) If the tube is placed in a uniform magnetic field, in what direction is the electron beam deflected?
(c) Calculate the magnitude of acceleration of an electron if the field strength is 3.65×10-4T.
downward
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64. Particle Orbits in a Uniform B
Tie a string to a rock and twirl it at constant speed in a circle over your head.
The tension of the string provides the centripetal force that keeps the rock moving in a circle.
The tension on the string always points to the center of the circle and creates a centripetal acceleration.
A particle with charge q and mass m moves with velocity v perpendicular to a uniform magnetic field B.
The particle will move in a circle with a constant speed v and the magnetic force F = qvB will keep the particle moving in a circle.
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65. Particle Orbits in Uniform B (2)
Recall centripetal acceleration
Newton;’s second law
So, for charged particle q in circular motion in magnetic field B
a = v2/r
F = m a
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66. Clicker Question
The figure shows the circular paths of two particles that travel at the same speed in a uniform magnetic field (directed into the page). One particle is a proton and the other is an electron. Which particle follows the smaller circle?
A) the electron
… for same speeds, r is proportional to m
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67. Example: Mass Spectrometer (1)
Suppose that B=80mT, V=1000V. A charged ion ( C) enters the chamber and strikes the detector at a distance x=1.6254m. What is the mass of the ion?
Key Idea: The uniform magnetic field forces the ion on a circular path and the ion’s mass can be related to the radius of the circular trajectory.
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68. Example: Mass Spectrometer (2)
Suppose that B=80mT, V=1000V. A charged ion ( C) enters the chamber and strikes the detector at a distance x=1.6254m. What is the mass of the ion?
From the figure: r=0.5x
Also need the velocity v after the ion is accelerated by the potential difference V
Blackboard calculation
Now solve for m
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3.4x10-25 kg

69. Example - Momentum of a Track
Calculate the momentum of this track
r = 2.3 m
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70. Example: Deuteron in Cyclotron
Suppose a cyclotron is operated at frequency f=12 MHz and has a dee radius of R=53cm. What is the magnitude of the magnetic field needed for deuterons to be accelerated in the cyclotron (m= kg)?
Key Idea: For a given frequency f, the magnetic field strength, B, required to accelerate the particle depends on the ratio m/q (or mass to charge):
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71. Special Clicker
Suppose a cyclotron is operated at frequency f=12 MHz and has a dee radius of R=53cm. What is the kinetic energy of the deuterons in this cyclotron when they travel on a circular trajectory with radius R (m= kg, B=1.57 T)?
A) J
B) J
C) J
D) J
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72. Forces and particles: Ex 2
A proton has a speed of 5.0 X 106 m/s and feels a force of 8.0 X 10-14N toward the west as it moves vertically upward.
Calculate the magnitude of the magnetic field.
Predict its direction
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73. vproton
Fwest
Earth
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74. Using right-hand rule: B must be towards geographic north
vproton
Fwest
Earth
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75. F = qvBsinq B = F/qvsinq = 8.0 X 10-14N
(1.6X10-19C)(5.0X106m/s)(sin90o)
B = 0.10 T
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76. Forces and particles: Ex 4
An electron travels at 2.0 X 107 m/s in a plane perpendicular to a T magnetic field. Describe its path.
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77. Path is circular (right-hand rule, palm positive) F = mv2 r qvB = mv2 r = mv qB
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78. r = mv qB r = (9. 1 X 10-31 kg)(2. 0 X 107 m/s ) (1. 6 X 10-19 C)(0
r = mv qB r = (9.1 X kg)(2.0 X 107 m/s ) (1.6 X C)(0.010 T) r = m
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