Magnetic Force on a ChargedParticle

• Magnets and Magnetic Fields– Magnets cause space to be modified in their vicinity,

forming a “magnetic field”.– The magnetic field caused by magnetic “poles” is

analogous to the electric field caused by electric “poles” or “charges”.

– The north pole is where the magnetic field lines leave the magnet, and the south pole is where they reenter.

– Magnetic field lines differ from electric field lines in that they are continuous loops with no beginning or end.

MagneticField, B

More Magnetic Fields

• Compasses– If they are allowed to select their own orientation,

magnets align so that the north pole points in the direction of the magnetic field.

– Compasses are magnets that can easily rotate so that they can align themselves to a magnetic field.

– The north pole of the compass points in the direction of the magnetic field.

Sample Problem

• A compass points to the Earth’s North Magnetic Pole (which is near the North Pole).

• Is the North Magnetic Pole the north pole of the Earth’s Magnetic Field?

Magnetic “Monopoles”

• Do not exist!– This is another way that magnetic fields differ

from electric fields.– Magnetic poles cannot be separated from

each other in the same way that electric poles (charges) can be.

Units of Magnetic Field

• Tesla (SI) – N/(C m/s) – N/(A m)• Gauss – 1 Tesla = 104 gauss

• Magnetic Force on Particles– Magnetic fields cause the existence of magnetic forces.– A magnetic force is exerted on a particle within a

magnetic field only if– the particle has a charge.– the charged particle is moving with at least a portion

of its velocity perpendicular to the magnetic field.

Magnetic Force on a ChargedParticle

• magnitude: F = qvBsinθ– q: charge in Coulombs– v: speed in meters/second– B: magnetic field in Tesla– q: angle between v and B

• direction: Right Hand Rule• FB = q v x B (This is a “vector cross product”)

The Right Hand rule to Determine aVector Cross Product

1. Align your hand along the first vector.2. Orient your wrist so that you can “cross” your hand

into the second vector.3. Your thumb gives you the direction of the third

vector (which is the result).

• Sample Problem:– Calculate the magnitude of force exerted on a 3.0 μC

charge moving north at 300,000 m/s in a magnetic field of 200 mT if the field is directed a)N, b)E, c)S, d)W

a) F = qvBsin= qvBsinθθ

F = 3x10F = 3x10-6-6C ·3x10C ·3x1055m/s ∙ 0.2T ∙ sin0m/s ∙ 0.2T ∙ sin0oo = 0 N= 0 N

b) F = qvBsin= qvBsinθθ

F = 3x10F = 3x10-6-6C ·3x10C ·3x1055m/s ∙ 0.2T ∙ sin90m/s ∙ 0.2T ∙ sin90oo = 0.18 N= 0.18 Nc) F = qvBsin= qvBsinθθ

F = 3x10F = 3x10-6-6C ·3x10C ·3x1055m/s ∙ 0.2T ∙ sin180m/s ∙ 0.2T ∙ sin180oo = 0 N= 0 Nd) F = qvBsin= qvBsinθθ

F = 3x10F = 3x10-6-6C ·3x10C ·3x1055m/s ∙ 0.2T ∙ sin270m/s ∙ 0.2T ∙ sin270oo = -0.18 N= -0.18 N

• Calculate the magnitude and direction of the magnetic force.

v = 300,000 m/s

B = 200 mTq = 3.0μC

34o

F = qvBsin= qvBsinθθF = 3x10= 3x10-6-6C ∙ 300000 m/s ∙ 0.2T ∙ sin(34C ∙ 300000 m/s ∙ 0.2T ∙ sin(34oo) )

F =0.101N upward

Magnetic forces…

• are always orthogonal (at right angles) to the plane established by the velocity and magnetic field vectors.

• can accelerate charged particles by changing their direction.

• can cause charged particles to move in circular or helical paths.

Magnetic forces cannot...

• change the speed or kinetic energy of charged particles.

Magnetic Forces on Charged Particles …

• …are centripetal.• Remember centripetal force is mv2/r.• For a charged particle moving perpendicular

to a magnetic field– F = qvB = mv2/r

• Radius of curvature of the particle– r = mv2/qvB = mv/qB

• A magnetic field of 2000 mT is directed to the south. A proton is traveling at 300,000 m/s to the west. What is the magnitude and direction of the force on the proton? What is the radius of curvature? Ignore gravitational effects.

90o

B = 2TB = 2T v = 300000m/sv = 300000m/s

F = qvBsin= qvBsinθθ

Sin(90Sin(90oo)) = 1

F = qvB= qvB F= 1.6 x 10F= 1.6 x 10-19-19C ∙300000m/s ∙2TC ∙300000m/s ∙2T = 9.6 x 10= 9.6 x 10-14-14 N N

Based on the right hand rule the force is coming out Based on the right hand rule the force is coming out of the pageof the page F = mvF = mv22/r/r

m = 1.67 x10m = 1.67 x10-27-27 Kg Kg

r = 1.67 x10r = 1.67 x10-27-27 Kg ∙ (30000m/s) Kg ∙ (30000m/s)2 2 / 9.6 x 10/ 9.6 x 10-14-14 N Nr= mvr= mv22/F/F

r =0.00157 mr =0.00157 m

• An electric field of 2000 N/C is directed to the south. A proton is traveling at 300,000 m/s to the west. What is the magnitude and direction of the force on the proton? Describe the path of the proton? Ignore gravitational effects.

E = 2000 N/CE = 2000 N/C

v = 300000m/sv = 300000m/s

+ + + + + + + +

- - - - - - - - - - -

+

vvxx constant in the x direction so F constant in the x direction so Fx netx net = 0 N = 0 N

FFyy = qE = qE FFyy = 1.67x10 = 1.67x10-19-19 C ∙ 2000 N/C C ∙ 2000 N/C = 3.2 x 10= 3.2 x 10-16-16 N N

F = mvF = mv22/r/rr = 1.67 x10r = 1.67 x10-27-27 Kg ∙ (30000m/s) Kg ∙ (30000m/s)2 2 / 3.2 x 10/ 3.2 x 10-16-16 N N

r= mvr= mv22/F/F

r =0.47 mr =0.47 m

Magnetic Force on Current-Carrying Wire

• F = I L B sinθ– I: current in Amps– L: wire length in meters– B: magnetic field in Tesla– θ: angle between current and field

• Affect moving charge– F = q v B sinθ– F = I L B sin θ– Hand rule is used to determine direction of this force.– They are also caused by moving charge

Sample Problem

• What is the force on a 100 m long wire bearing a 30 A current flowing north if the wire is in a westward-directed magnetic field of 400 mT?

F = I L B sinθ = 30A (100m)(400)(10-3)T (sin90o) = 1200N

Right Hand Rule For straight currents1. Curve your fingers2. Place your thumb (which is

presumably pretty straight) in direction of current.

3. Curved fingers represent curve of magnetic field.

4. Field vector at any point is tangent to field line.

Magnitude of Magnetic Field for Straight Currents.

• B = μoI / (2πr)

– μo: 4π × 10-7 T m / A

• magnetic permeability of free space

– I: current (A)

– r: radial distance from center of wire (m)

• What is the magnitude and direction of the magnetic field at point P, which is 3.0 m away from a wire bearing a 13.0 Amp current?

B = μB = μooI / (2I / (2ππr)r)

B = (4B = (4ππ x 10 x 10-7-7 T ∙ m/A ∙ 13A) / (2 ∙ T ∙ m/A ∙ 13A) / (2 ∙ ππ ∙ 3.0m) ∙ 3.0m)

B = 8.67 x 10B = 8.67 x 10-7 -7 TT

• Principle of Superposition– When there are two or more currents forming a

magnetic field, calculate B due to each current separately and then add them together using vector addition.

• What is the magnitude and direction of the electric field at point P if there are two wires producing a magnetic field at this point?

I1= 10.0 A

I2= 13.0 AP ●

4.0m

3.0m

B = μB = μooI / (2I / (2ππr)r)

BB11 = (4 = (4ππx10x10-7-7)(10A)/(2∙)(10A)/(2∙π∙π∙4m) = 5x104m) = 5x10-7-7 T T

BB22 = (4 = (4ππx10x10-7-7)(13A)/(2∙)(13A)/(2∙π∙π∙3m) = 8.67x103m) = 8.67x10-7-7 T T

BB11 + B + B22 = 13.67X10 = 13.67X10-7-7 T T(why do the B’s add?) (use RHR)(why do the B’s add?) (use RHR)

What would B1 + B2 be if the currents were in the same What would B1 + B2 be if the currents were in the same direction? What would be the direction – in or out of the direction? What would be the direction – in or out of the paper?paper?

XX

XX

XX

XX

BB11 - B - B22 = 3.67X10 = 3.67X10-7-7 T T

• In the 4th Grade• You learned that coils

with current in them make magnetic fields.

• The iron nail was not necessary to

• cause the field; it merely intensified it.

• Solenoid– A solenoid is a coil of wire.– When current runs through the wire, it causes the coil

to become an “electromagnet”.– Air-core solenoids have nothing inside of them.– Iron-core solenoids are filled with iron to intensify the

magnetic field.

• Right Hand Rule for magnetic fields around curved wires1. Curve your fingers.2. Place them along wire loop

so that your fingers point in direction of current.

3. Your thumb gives the direction of the magnetic field in the center of the loop, where it is straight.

4. Field lines curve around and make complete loops.

I

B

XX

• What is the direction of the magnetic field produced by the current I at A? At B?

I

BXXA

• Magnetic Flux– The product of magnetic field and area.– Can be thought of as a total magnetic “effect” on a coil

of wire of a given area.

– Max flux – when the area is perpendicular to the magnetic field

– or when the normal line of the surface of the area is parallel to the magnetic field

B

A

– Minimum flux – when the area is parallel to the magnetic field

– or the normal line to the surface of the area is perpendicular to the magnetic field

B

A

– Intermediate flux – when the area is neither parallel or perpendicular to the magnetic field

– or the normal line to the surface of the area is neither perpendicular nor parallel to the magnetic field

B

A

• Magnetic Flux– ΦB = B A cosθ

– ΦB : magnetic flux in Webers (Wb) or (T m∙ 2)

– B: magnetic field in Tesla– A: area in meters2.– θ: the angle between the area and the– magnetic field.– ΦB = B×A

• Calculate the magnetic flux through a rectangular wire frame 3.0 m long and 2.0 m wide if the magnetic field through the frame is 4.2 mT.

• a) Assume that the magnetic field is perpendicular to the area vector.

• b) Assume that the magnetic field is parallel to the area vector.

• c) Assume that the angle between the magnetic field and the area vector is 30o.

A = 3.0m x 2.0m = 6 m2

B = 0.0042 T

A = 3.0m x 2.0m = 6 m2

Induced Electric Potential(Faraday’s Law of Induction)

• A system will respond so as to oppose changes in magnetic flux.

• A change in magnetic flux will be partially offset by an induced magnetic field whenever possible.

• Changing the magnetic flux through a wire loop causes current to flow in the loop.

• This is because changing magnetic flux induces an electric potential.

Faraday’s Law of Induction

• ε = - N(ΔΦB /Δt) = -N(Δ[BAcosθ]/Δt)

– ε: induced potential (V)– N: # loops– ΦB : magnetic flux (Webers, Wb)

– t: time (s)• To generate voltage, change B or change A or change θ

Sample Problem

• A single coil of radius 0.25 m is in a 100 mT magnetic field such that the flux is maximum. At time t = 1.0 seconds, field increases at a uniform rate so that at 11 seconds, it has a value of 600 mT. At time t = 11 seconds, the field stops increasing. What is the induced potential –A) at t = 0.5 seconds?B) at t = 3.0 seconds?C) at t = 12 seconds?