-Magnetic Force on a Charged Particle-Magnetic Force on a Current-Carrying Wire-Torque on a Current-Carrying Loop

AP Physics CMrs. Coyle

Fermi Lab, Chicago Illinois Circumference 6.3km

Mass Spectrometer

DC Motor

• A magnetic field can exert a force on a charged particle that moves in it.

Three ways we talk about “magnetic field”.

• Magnetic Field: Regions surrounding a magnet where another magnet or a moving electric charge will feel a force of attraction or repulsion.

• Magnetic Field Lines: exit the north pole and enter the south.

• Magnetic Field Strength, B

• Vector, Unit: Tesla, T• Named after Nikola Tesla

Poles• Law of Poles: Like poles repel, unlike poles attract.• The force between two poles varies as the inverse

square of the distance between them.• A single pole (monopole) has not been isolated.

Force on a Charged Particle Moving in a Magnetic Field

-Magnetic fields only exert forces on moving charged particles or other magnets. 

• F = |q|(v x B) = |q|vB sin

• vector cross-product

is the angle between v and B• F=0 for =0 or 1800

Remember: Cross Product Using Determinants

or

ˆ ˆ ˆ

ˆ ˆ ˆy z x yx zx y z

y z x yx zx y z

A A A AA AA A A

B B B BB BB B B

i j k

A B i j k

ˆ ˆ ˆy z z y x z z x x y y xA B A B A B A B A B A B A B i j k

Remember: Properties of Cross Remember: Properties of Cross Products of Unit VectorsProducts of Unit Vectors

ixi=0

jxj=0

kxk=0

ixj=k

jxk=i

kxi=j

To find the direction of the force use the

Right Hand Rule

For a positive test charge:

Thumb v

Fingers B

Out of palm F

The force is always perpendicular to the vB plane

For a negative particle the F is opposite to what it would be for a positive particle

(use left hand)

Alternate Rule: Right Hand Curl Rule

•Curl fingers from v to B F = q(v x B)

•F is in the direction of the thumb

•Similarly used for the direction of torque (=r x F)

Graphical Representation of the Magnetic Field Vector (Strength), B

x field lines pointing into the page

● field lines pointing out of the page 

Question• Find the direction of the magnetic force

acting on the +charged particle entering the magnetic field with a velocity v perpendicular to B.

x x x x x

x x x x x

V x x x x x

x x x x x

Answer: Upwards

What motion will the particle in the previous example undergo (particle entered the B-

field in a direction perpendicular to B?• Circular Motion

• Magnetic force will represent the centripetal force

• http://online.cctt.org/physicslab/content/applets/JavaPhysMath/java/partmagn/index.html

Examples:

http://physicslearning.colorado.edu/PiraHome/PhysicsDrawings.htm

Does the magnetic force do work?

• F is always perpendicular to the displacement

• F can change the direction of v not the magnitude

• F cannot do work, cannot change KE

Mass Spectrometer U.S. Department of Energy

http://doegenomestolife.org

Motion of Alpha- Beta-Gamma Particles in a Magnetic Field

1) Alpha particles, positive helium nuclei, charge +2e

2) Gamma rays, (no charge) electromagnetic radiation

3) Beta particles, electrons charge -1e

Problem 1A proton is accelerated through a constant electric field

(parallel plates) and acquires kinetic energy of 4eV.

It enters perpendicularly to the 2T field of a detector as shown.

Charge of a proton=1.6x10-19 C, mass of proton=1.67x10-27 kg, ignore gravity.

a) Draw the path of the positive charge as it enters the magnetic field.

Problem 1 cont’db) Calculate the force acting on the charge due to the

magnetic field.

Charge of a proton=1.6x10-19 C , mass of proton=1.67x10-27 kg

KE= 4eV B= 2T

Ans: v= 2.77x104 m/s, F= 8.86x10-15 N

Problem 1 cont’dc) Calculate the distance on the detector where the particle will

land (radius of the circular path). Ignore gravity.

Charge of a proton=1.6x10-19 C, mass of proton=1.67x10-27 kg KE= 4eV B= 2T

Ans: 1.45x10-4 m

Large Hadron ColliderCERN (Conseil Européen pour la Recherche

Nucléaire) Switzerland- France 2008

Circumference: 27km

• Particle Accelerators use electric and magnetic fields to accelerate charged particles.

• Cyclotron Applet

Magnetic Force on a Current Carrying WireF = I (L x B) = I L B sin

What is the direction of the magnetic force acting on this wire?

F

• I is the current

• L is a vector of magnitude of the length of the wire and direction that of the current

•is the angle formed between I and B

The force acting on a wire of arbitrary shape is the same as if it were a

straight wire with the same ends• The total force is:

I sin

bd ILB θ

aF s B

What is the net force acting on this current-carrying loop?

I bd

aF s B

Which way will a loop turn?

Forces on a Current-Carrying Loop

For B as shown:F 1 = F3 = 0

F 2 = F4 = IaB

Electric DC Motor

(Fendt Applet)

Top View

(I ) (I )

I

I

2 42 2 2 2

, where A is the area of the loop

b b b bτ F F aB aB

τ abB

τ AB

F 2 = F4 = IaB

Torque acting on the loop:

Torque Acting on the Current-Carrying Loop

Torque Acting on the Current-Carrying Loop

sin

sin

In general:

, where A is the area of the loop

=IA x B

Magnetic Dipole Moment

τ IabB θ

τ IAB θ

τ

μ IA

Magnetic Force on a Current Carrying Wire

• Lorentz Force- Magnetic Force on a Current Carrying Wire (Fendt Applet)

DC Motor

DC Motor

Problem #9

A proton moves with a velocity of v=(2i-4j+k)m/s

in a region in which the magnetic field is

B=(i+2j-3k)T. What is the magnitude of the magnetic force this charge experiences?

Ans: 2.34x10-18 N

Problem #13

A wire 2.80 m in length carries a current of 5.00 A in a region where a uniform magnetic field has a magnitude of 0.390 T. Calculate the magnitude of the magnetic force on the wire assuming the angle between the magnetic field and the current is (a) 60.0°, (b) 90.0°, (c) 120°.

Ans: a)4.73N, b)5.46N, c)4.73N

Problem # 21

A small bar magnet is suspended in a uniform 0.250-T magnetic field. The maximum torque experienced by the bar magnet is 4.60 × 10–3 N · m. Calculate the magnetic moment of the bar magnet.

Ans: 18.4mA m2

Problem #54

A 0.200-kg metal rod carrying a current of 10.0 A glides on two horizontal rails 0.500 m apart. What vertical magnetic field is required to keep the rod moving at a constant speed if the coefficient of kinetic friction between the rod and rails is 0.100?

Ans: F=0.196N, B=0.039T