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29.1 MAGNETIC FIELDS AND FORCES29.2 MAGNETIC FORCE ACTING ON A CURRENT-CARRYING CONDUCTOR29.4 MOTION OF A CHARGED PARTICLE IN A UNIFORM MAGNETIC FIELD29.5 APPLICATIONS INVOLVING CHARGED PARTICLES MOVING IN AMAGNETIC FIELD

CHAPTER 29 MAGNETIC FIELDS

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MagnetismSince 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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Magnetic Field LinesWe 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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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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Force Due to Magnetic FieldThe force exerted on a charged particle by a magnetic field is given

by the vector cross product:

F = q v B sin

F = force (vector)

q = charge on the particle (scalar)

v = velocity of the particle relative to field (vector)

B = magnetic field (vector)

F = q v 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

where is angle between q v and B vectors.

Bq

V

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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.

sinqvBF

BvqvBF for max

BvF //for 0

The force is greatest when the velocity v is perpendicular to the B field. The deflection decreases to zero for parallel motion.

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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• 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 B

q

V

A magnetic field of one tesla is very powerful magnetic field. Sometimes it may be convenient to use the gauss, which is equal to 10 -4 tesla.

1 N = 1 C (m / s) (T)F = q v B sin

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 T = 1 N

C (m / s)

1 N A m

=

. If a charge moves parallel to a magnetic field, there is no magnetic force on it, since sin 0° = 0.

• 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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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.

or sinsin

FB F qvB

qv

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).

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).

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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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Right Hand Rule• Direction of force on a positive charge given by the right hand

rule.

BvqF

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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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1. Left2. Up3. Down4. Into the page5. Out of the page

An electron moves perpendicular to a magnetic field. What is the direction of ?B

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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 = 0.0015 N directed out of the page.

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ExampleA 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 10-19 C.

v = 3 x 104 m/s, WestB = 500 Gauss * 1 Tesla/10,000 Gauss = .05 T, South

magnitude:Fmagnetic = q v B sin(θ)

direction: right hand rulemagnitude: F = (1.6 x 10-19 C) (3 x 104 m/s) * (.05 T) * sin(90o) = 2.4 x 10-16 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 10-16 Nt / 1.67 x 10-27 kg = 1.44 x 1011 m/s2.

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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?

v sin fv

300

B

v

FDraw a rough sketch.

q = 2 x 10-9 C v = 5 x 104 m/s B = 3 x 10-3 T q = 300

Using right-hand rule, the force is seen to be upward.

Resultant Magnetic Force: F = 1.50 x 10-7 N, upward Resultant Magnetic Force: F = 1.50 x 10-7 N, upward

B

Example

Fmagnetic = q v B sin(θ)

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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 10-27 kg, 1 eV = 1.6 x 10-19 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

N 10 6.1 1 101.2 103.2 10 1.6 sin qVB 15- 3- 719- FB

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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

· · ·· · · ·· · · ·· · · ·

·

What is the direction of the force F on the charge in each of the examples described below?

What is the direction of the force F on the charge in each of the examples described below?

-v

-v

+

v

v+

UpF

LeftF

F RightUp

F

negative q

negative q

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Direction of Magnetic Force on Moving Charges

Velocity B Forceout of page right upout of page left down

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1) Up 2) Down 3) Right 4) Left 5) Zero

out of page up

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Velocity B Forceout of page right upout of page left downout of page up out of page down right

left

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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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e(1.0107 ms )(0.5T)sin90

810 13N

Direction RHR

e(1.0107 ms )(0.5T)sin90

810 13N

Direction RHR

^

50.0 iTB

jv sm ˆ100.1 7

BvqFm

BvqFm

sinBvqFm

kjv s

m ˆ2

1ˆ2

1100.1 7

iTB ˆ50.0

sinBvqFm

kNFm ˆ108 13 kjNFm ˆˆ108

21

2113

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A) negative z direction B) positive y direction C) positive z direction

D) negative x 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?

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A) positive y direction B) positive z direction C) the force is 0

D) negative x 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?

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A) positive y direction B) the force is 0 C) negative z direction

D) negative x direction F = q v B sin(180) = 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?

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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

B

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.

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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

1 2

v = 75 m/sq = +25 mC

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.

Particle is moving straight upwards then veers to the right.

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What is the direction of the magnetic field in region 1?1) up2) down3) left4) right5) into page6) out of page

1 2

v = 75 m/sq = +25 mC

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.

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Motion of a Charge in a Magnetic FieldThe ’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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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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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:

Circulating Charged Particles

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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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Circulating Charged Particles

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:

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.

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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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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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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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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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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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.

FB = q v x B

FB = i L x B

Recall the force equation that defined the B-field in terms of the force acting a single moving charge.

This equation is almost equivalent to the force on a wire equation

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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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 Bto each segment. The net force on the wire is the vector sum of the forces on each of the segments.

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Force on a Current-Carrying Wire

F = ILB

In a wire carrying current the charge carriers are moving at the drift velocity vd (on average),

what is the force on the wire?

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What is the direction?

Vector form:

F = I LBwhere L is a vector along the wire segment in the direction of positive current

L

Right hand rule gives direction

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Force: Ex. 2

A loop of wire carries 0.245 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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F = IlBsinqF = IlBsin90o

F = IlB(1)F = IlBB = F

Il

B = 3.48 X 10-2 N = 1.42 T (0.245 A)(0.100m)

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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 10-19 Coul; B = .05 Teslas (500 G)m = 1.67 x 10-27 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 10-

19 Coul)*(0.05 T) = .065 m = 6.5 cm.

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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.

F F

original path subsequent path

point ofvelocity reversal

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A Neon ion, q = 1.6 x 10-19 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?

qBRm

v

1000 V/m

0.5 T

Ev

B

v = 2000 m/s

m = 2.91 x 10-24 kgm = 2.91 x 10-24 kg

+q

R

Ev

B

+- x x x x x x x x

Photographic plate

m

slitx 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

Example

m v

R = q B

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What is the speed of the particle when it leaves chamber 2?

1) v2 < v1

2) v2 = v1

3) v2 > v1

1 2

v = 75 m/sq = +25 mC

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.

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Magnetic force is always perpendicular to velocity, so it changes direction, not speed of particle.

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Compare the magnitude of the magnetic field in chambers 1 and 2

1) B1 > B2

2) B1 = B2.

3) B1 < B2

1 2

v = 75 m/sq = +25 mC

Larger B, greater force, smaller R

qBmv

R

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.

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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

1 2

v = 75 m/sq = ?? mC

qBmv

R

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.

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1. Which angle do you use to determine the magnitude of the force on the charged particle? (q1, q2 , either one).

2. Below I have drawn the trajectory for two charged particles traveling through a magnetic field. Is particle 1 positive or negatively charged?

3. If particles 1 and 2 have the same mass and velocity, which has the largest charge?

B

Vq1

q2

1

2

Either one sin(q) = sin(180-q)

Positive

Particle 1

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II

B into the p age

wire

F I Bo n w ire

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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Magnetic interactions:1. A moving charge or a

current creates a magnetic field in the surrounding space

2. The magnitude field exerts a force on any other moving charge or current that is present in the field

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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?

31 10 1

19

(9.11 10 )(1.54 10 )

1.60 100.0877

mB

q

kg s

CT

Example

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Velocity Selector

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.

v

E

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) Right38

FE would be up so FB must be down.

FE

FB

Electric force is down, so need magnetic force up. By RHR, B must be into page

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A lithium ion, q = +1.6 x 10-16 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

Ev

B

E = vB

E = (1.5 x 106 m/s)(20 x 10-3 T); E = 3.00 x 104 V/mE = 3.00 x 104 V/m

Example.

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Let’s find a relationship between q, B, and E if there is no deflection at all:

FB = FE

q v B = q E v =

Fnet = 0E

B

Need force up if proton is to be undeflected Put magnetic field into page

Then

Fy may

Fe y Fm y may qE qBv may

We want ay = 0

qE qBv

E Bv

jBvqBvqFm ˆ

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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 9.109410-31 kg while the elementary charge is 1.602210-19 C.• (a) Calculate the velocity of the electrons in the beam after leaving the electron gun.

m/s 1092.6 implies 6221

veVmv

qVUK

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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.

F ma qvB

a qvB

m

1.602210 19 C 6.92106 m/s 3.6510 4 T 9.1094 10 31 kg

4.44 1014 m/s2

downward

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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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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

Bvqrv

m 2

a = v2/r

F = m a

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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

r mv

qB… for same speeds, r is proportional to m

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Example: Mass Spectrometer (1)• Suppose that B=80mT, V=1000V. A

charged ion (1.6022 10-19C) 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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Example: Mass Spectrometer (2)• Suppose that B=80mT,

V=1000V. A charged ion (1.6022 10-19C) 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 eVmv 2

21

evBr

mv

2

Now solve for m

3.4x10-25 kg

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Example - Momentum of a Track• Calculate the momentum of this track

r = 2.3 m

36.0100.31067.1

108.1

m/s kg 108.1

/

827

19

19

2

mcmv

cv

erBmv

evBrmv

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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=3.34 10-27kg)?

• 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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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=3.34 10-27kg, B=1.57 T)?

A) 0.9 10-14 J B) 8.47 10-13 J C) 2.7 10-12 J D) 3.74 10-13 J

J 107.2

m/s 1099.3 implies

12221

7

mvK

mRqB

vqBmv

r

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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.

a. Calculate the magnitude of the magnetic field.

b. Predict its direction

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Earth

Fwest

vproton

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Using right-hand rule:B must be towards

geographic north

Earth

vproton

Fwest

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F = qvBsinqB = F/qvsinqB= 8.0 X 10-14N (1.6X10-19C)(5.0X106m/s)(sin90o)

= 0.10 B T

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Forces and particles: Ex 4

An electron travels at 2.0 X 107 m/s in a plane perpendicular to a 0.010-T magnetic field. Describe its path.

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Path is circular (right-hand rule, palm positive)F = mv2

r

qvB = mv2

r

r = mv qB

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r = mv qB

r = (9.1 X 10-31 kg)(2.0 X 107 m/s )(1.6 X 10-19 C)(0.010 T)

r = 0.011 m