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Magnetic Forces

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Magnetic Forces. Forces in Magnetism. The existence of magnetic fields is known because of their affects on moving charges. What is magnetic force (F B )? How does it differ from electric force (F E )? What is known about the forces acting on charged bodies in motion through a magnetic field? - PowerPoint PPT Presentation

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Page 1: Magnetic Forces

Magnetic Forces

Page 2: Magnetic Forces

Forces in Magnetism The existence of magnetic fields is known

because of their affects on moving charges. What is magnetic force (FB)? How does it differ from electric force (FE)? What is known about the forces acting on

charged bodies in motion through a magnetic field?• Magnitude of the force is proportional to the

component of the charge’s velocity that is perpendicular to the magnetic field.

• Direction of the force is perpendicular to the component of the charge’s velocity perpendicular to the magnetic field(B).

Page 3: Magnetic Forces

Magnetic Force (Lorentz Force)

FB = |q|vB sinθ Because the magnetic force is always

perpendicular to the component of the charge’s velocity perpendicular to the magnetic field, it cannot change its speed.

Force is maximum when the charge is moving perpendicular to the magnetic field ( = 90).

The force is zero if the charge’s velocity is in the same direction as the magnetic field ( = 0).

Also, if the speed is not changing, KE will be constant as well.

Page 4: Magnetic Forces

Example #1 A positively charged particle traveling at

7.5 x 105 meters per second enters a uniform magnetic field perpendicular to the lines of force. While in the 4.0 x 10-2 tesla magnetic field, a net force of 9.6 x 10-15 newton acts on the particle. What is the magnitude of the charge on the particle?

CT

N

vB

Fq B 19

25

15

102.3)100.4)(105.7(

106.9

Page 5: Magnetic Forces

What is the magnetic field (B)? The magnetic field is a force field just like electric and

gravitational fields. It is a vector quantity. Hence, it has both magnitude and direction.

Magnetic fields are similar to electric fields in that the field intensity is directly proportional to the force and inversely related to the charge.

E = FE/q

B = FB/(|q|v)

Units for B: N•s/C•m = 1 Tesla

Page 6: Magnetic Forces

Right Hand Rules Right hand rule is used to determine the

relationship between the magnetic field, the velocity of a positively charged particle and the resulting force it experiences.

Page 7: Magnetic Forces

Right Hand Rules

#1 #2 #3

FB = |q|v x B

Page 8: Magnetic Forces

The Lorentz Force Equation & RHR

What is the direction of force on the particle by the magnetic field?

a. Right b. Left c. Up d. Downe. Into the page f. Out of the Page

V

+Uniform Bθ

vsinθ

q

FB = qvB sinθ

Note: Only the component of velocity perpendicular to the magnetic field (vsin) will contribute to the force.

Page 9: Magnetic Forces

Right Hand Rule – What is the Force?

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

What is the direction of the magnetic force on the charge?

a) Down b) Up c) Right d)Left

Page 10: Magnetic Forces

Right Hand Rule – What is the Charge?

Particle 1:a. Positiveb. Negativec. Neutral

Particle 2:a. Positiveb. Negativec. Neutral

Particle 3:a. Positiveb. Negativec. Neutral

Page 11: Magnetic Forces

Right Hand Rule – What is the Direction of B

What is the direction of the magnetic field in each chamber?

a. Upb. Downc. Leftd. Righte. Into Pagef. Out of Page

12 3

4

What is the speed of the particle when it leaves chamber 4?

a. v/2 b. -vc. v d. 2v

Since the magnetic force is always perpendicular to the velocity, it cannot do any work and change its KE.

Page 12: Magnetic Forces

Two protons are launched into a magnetic field with the same speed as shown. What is the difference in magnitude of the magnetic force on each particle?

a. F1 < F2 b. F1 = F2 c. F1 > F2

Example 2: Lorentz Force

x x x x

x x x x

x x x x

v2

+

+ v1

F = qv x B = qvBsinθSince the angle between B and the particles is 90o in both cases, F1 = F2.

How does the kinetic energy change once the particle is in the B field?

a. Increase b. Decrease c. Stays the Same

Since the magnetic force is always perpendicular to the velocity, it cannot do any work and change its KE.

Page 13: Magnetic Forces

Trajectory of a Charge in a Constant Magnetic Field What path will a charge take when it enters a constant

magnetic field with a velocity v as shown below?

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 Since the force is always perpendicular to the v and B, the

particle will travel in a circle Hence, the force is a centripetal force.

Page 14: Magnetic Forces

Radius of Circular Orbit

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 radius of the circular orbit?

Lorentz Force: F = qv x B

Centripetal Acc: ac = v2/R

Newton’s Second Law: F = mac

qvB = mv2/R

R = mv/qB

v

+

Fc

R

Page 15: Magnetic Forces

Example #2 A particle with a charge of 5.0 x 10-6 C traveling

at 7.5 x 105 meters per second enters a uniform magnetic field perpendicular to the lines of force. The particle then began to move in a circular path 0.30 meters in diameter due to a net force of 1.5 x 10-10 newtons. What is the mass of the particle?

TCsm

N

qv

FB B 11

65

10

100.4)100.5)(/105.7(

105.1

kgsm

mT

v

BqRm 23

5

611

100.4)/105.7(

)15.0)(100.5)(100.4(

Page 16: Magnetic Forces

Earth’s Magnetosphere

Magnetic field of Earth’s atmosphere protects us from charged particles streaming from Sun (solar wind)

Page 17: Magnetic Forces

Aurora

Charged particles can enter atmosphere at magnetic poles, causing an aurora

Page 18: Magnetic Forces

Crossed Fields in the CRT

How do we make a charged particle go straight if the magnetic field is going to make it go in circles? Use a velocity selector that

incorporates the use of electric and magnetic fields.

Applications for a velocity selector: Cathode ray tubes (TV, Computer

monitor)

Page 19: Magnetic Forces

Crossed Fields

E and B fields are balanced to control the trajectory of the charged particle.

FB = FE

Velocity Selector

qvB = qE

v = E/B

B into page

x x x x

x x x x

x x x x

+ + + + +

- - - - -E

- v - v

Phosphor Coated Screen

- v

FB

FE

Page 20: Magnetic Forces

Force on a Current Carrying WireFB = |q|v x B = qvB sinθ (1)

Lets assume that the charge q travels through the wire in time t.

FB = (q)vBsinθ

When t is factored in, we obtain:

FB = (q/t)(vt) Bsinθ (2)

Where: q/t = I (current) vt = L (length of wire)

Equation (2) therefore reduces to:

FB = ILB sinθ

Page 21: Magnetic Forces

Examples #2 & #3 A wire 0.30 m long carrying a current of 9.0

A is at right angles to a uniform magnetic field. The force on the wire is 0.40 N. What is the strength of the magnetic field?

TmA

N

IL

FB B 15.0

)30.0)(0.9(

4.0

AmT

N

BL

FI 3100.6

)650)(46(.

)8.1(

A wire 650 m long is in a 0.46 T magnetic field. A 1.8 N force acts on the wire. What current is in the wire?

Page 22: Magnetic Forces

Torque on a Current Carrying Coil (Electric Motors/Galv.)

= F•r

Page 23: Magnetic Forces

Torque on a Current Carrying Coil (cont.)

B

-F

F

xI

x

F

-F

Zero Torque

w • x

-F

F

Max Torque

Direction of Rotation

Axis of Rotation

Page 24: Magnetic Forces

Torque on a Current Carrying Coil (cont.) At zero torque, the magnetic field of the loop of

current carrying wire is aligned with that of the magnet.

At maximum torque, the magnetic field of the loop of current carrying wire is at 90o.

The net force on the loop is the vector sum of all of the forces acting on all of the sides.

When a loop with current is placed in a magnetic field, the loop will rotate such that its normal becomes aligned with the externally applied magnetic field.

Page 25: Magnetic Forces

Torque on a Current Carrying Coil (cont.)

What is the contribution of forces from the two shorter sides (w)?

F = IwB sin (90o – )

Note 1: is the angle that the normal to the wire makes with the direction of the magnetic field.

Note 2: Due to symmetry, the forces on the two shorter sides will cancel each other out (Use RHR #1).

L

w

IAxis of rotation

X X X X

X X X X

Page 26: Magnetic Forces

Torque on a Current Carrying Coil (cont.) What is the contribution of torque from the two longer

sides (L)?

F = BIL for each side since L is always perpendicular to B.

The magnitude of the torque due to these forces is:

= BIL (½w sin) + BIL (½w sin) = BILw sin (1)

Note: Since Lw = the area of the loop (A), (1) reduces to:

= IAB sin

For a winding with N turns, this formula can be rewritten:

= NIAB sin

Page 27: Magnetic Forces

DC Motor

DC Electric Motor

Armature – Part of the motor that spins that contains windings and an iron core.

External Magnetic Field – Electromagnet or permanent magnet that provides an attractive and repulsive force to drive armature.

Split Ring Commutator – Brushes and split ring that provide the electrical connection to the armature from the external electrical source.

Page 28: Magnetic Forces

Key Ideas Lorentz Force: A charge moving perpendicular to

a magnetic field will experience a force. Charged particles moving perpendicular to a

magnetic field will travel in a circular orbit. The magnetic force does not change the kinetic

energy of a moving charged particle – only direction.

The magnetic field (B) is a vector quantity with the unit of Tesla

Use right hand rules to determine the relationship between the magnetic field, the velocity of a positively charged particle and the resulting force it experiences.