b magnetism power point

8/4/2019 b Magnetism Power Point

http://slidepdf.com/reader/full/b-magnetism-power-point 1/73

MagnetismTuesday, February 22, 2011



Announcements



Magnetic Fields

s Magnets cause space to be modified in their vicinity, forming a “magnetic field”.

s

The magnetic field caused by magnetic“poles” is analogous to the electric fieldcaused by electric “poles” or “charges”.

s The north pole is where the magnetic field

lines leave the magnet, and the south pole iswhere they reenter.s Magnetic field lines differ from electric field

lines in that they are continuous loops with no

beginning or end.



Magnetic

Field,B



More Magnetic Fields



Compasses

s If they are allowed to select their own

orientation, magnets align so that the north

pole points in the direction of the magneticfield.

s Compasses are magnets that can easily

rotate so that they can align themselves to a

magnetic field.s The north pole of the compass points in the

direction of the magnetic field.



Sample Problem

s A compass points to

the Earth’s South

Magnetic Pole (whichis near the North

Pole).

s Is the North MagneticPole the north pole of

the Earth’s Magnetic

Field?



Magnetic “Monopoles”

s 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

s Tesla (SI)

– N/(C m/s)

– N/(A m)

s Gauss

– 1 Tesla = 104 gauss



Magnetic Force on Particles

s Magnetic fields cause the existence of

magnetic forces.

s A magnetic force is exerted on aparticle within a magnetic field only if

– the particle has a charge.

– the charged particle is moving with at leasta portion of its velocity perpendicular to the

magnetic field.



Magnetic Force on a Charged

Particles magnitude: F = qvBsinθ

– q: charge in Coulombs

– v: speed in meters/second – B: magnetic field in Tesla

θ: angle between v and B

s

direction: Right Hand Rules F

B= q v x B (This is a “vector cross

product”)



The Right Hand rule to Determine a

Vector Cross Product1. Align your thumb along the velocity vector.

2. Orient your index finger so that it points in

the direction of the magnetic field.

3. Your palm faces the direction of the force

vector (which is the result).



Sample Problem

Calculate the magnitude force exerted ona 3.0 µC charge moving north at 300,000

m/s in a magnetic field of 200 mT if thefield is directed

a) North.

b) South.c) East.

d) West.



Sample Problem

s Calculate the magnitude and direction

of the magnetic force.

v = 300,000 m/s

B = 200 mT

q = 3.0µC

34o



Wednesday, February 23, 2011

Motion of Charged Particles in

Magnetic Field



Magnetic forces…

s are always orthogonal (at right angles)

to the plane established by the velocity

and magnetic field vectors.s can accelerate charged particles by

changing their direction.

s can cause charged particles to move incircular or helical paths.



Magnetic forces cannot...

s change the speed or kinetic energy of

charged particles.

s do work on charged particles.



Magnetic Forces…

s …are centripetal .

s Remember that centripetal acceleration

is v2/r.

s Remember centripetal force is therefore

mv2/r.



Magnetic Forces are Centripetal

B

F

V

F

V

F

V

F

ΣF = ma

FB = Fc

qvBsinθ = mv2/rqB = mv/r

q/m = v/(rB)

V



Sample Problem

What is the orbital radius of a proton moving at

20,000 m/s perpendicular to a 40 T magneticfield?

l bl



Sample Problem

What must be the speed of an electron if it is to

have the same orbital radius as the proton inthe magnetic field described in the previous

problem?



Thursday, February 24th, 2011

Electric and Magnetic Fields

Together At Last

S l P bl



Sample Problem

An electric field of 2000 N/C is directed to the

south. A proton is traveling at 300,000 m/s tothe west. What is the magnitude and direction of

the force on the proton? Describe the path of

the proton? Ignore gravitational effects.



Sample Problem A magnetic field of 2000 mT is directed to the

south. A proton is traveling at 300,000 m/s tothe west. What is the magnitude and direction of

the force on the proton? Describe the path of

the proton? Ignore gravitational effects.



Sample Problem

s Calculate the force and describe thepath of this electron.

E = 2000 N/C

e-300,000 m/s



Sample Problem

s Calculate the force and describe thepath of this electron.

e-

300,000 m/s

B = 2000 mT



Sample problems How would you arrange a magnetic field and

an electric field so that a charged particle of velocity v would pass straight through without

deflection?



Electric and Magnetic Fields

Together

E

B

v = E/B

e-

S l P bl



Sample Problem

It is found that protons traveling at 20,000 m/s

pass undeflected through the velocity filter below. What is the magnitude and direction of

the magnetic field between the plates?

400 V

e20,000 m/s

0.02 m



Friday, February 25, 2011

Magnetic Force on Current

Carrying Wires



Magnetic Force on Current-

Carrying Wires F = I L B sinθ

– I: current in Amps

– L: length in meters

– B: magnetic field in Tesla

θ: angle between current and field

S l P bl



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

magnetic field (directed into the page) of 400

mT?

S l P bl



Sample Problem

What is the magnetic field strength if the currentin the wire is 15 A and the force is downwardand has a magnitude of 40 N/m? What is thedirection of the current?



Magnetic Fields…

s Affect moving charge

– F = qvBsinθ

– F = ILBsinθ

– Hand rule is used to determine direction of

this force.

s Caused by moving charge!



Magnetic Field for

Long Straight Wires 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)

Ri h H d R l



Right Hand Rule

for straight currents

1. Curve your fingers

2. Place your thumb (which ispresumably pretty straight)in direction of current.

3. Curved fingers representcurve of magnetic field.

4. Field vector at any point is

tangent to field line.

i

r • ×



For

straightcurrents

I

Sample Problem



Sample Problems 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?

I = 13.0 A

P

3.0 m



Monday, February 28, 2011

Superposition in Magnetic Fields



Principle of Superposition

s When there are two or more currents

forming a magnetic field, calculate B

due to each current separately and thenadd them together using vector

addition.

Sample Problem



Sample Problems What is the magnitude and direction of the

magnetic field at point P if there are two wires

producing a magnetic field at this point?

I = 13.0 A

P

3.0 m

I = 10.0 A

4.0 m

Sample Problem



Sample Problem

s Where is the magnetic field zero?

I = 13.0 A

7.0 m

I = 10.0 A

Sample Problem not in notes



Sample Problem – not in notess What is the magnitude and direction of the

force exerted on a 100 m long wire that

passes through point P which bears a currentof 50 amps in the same direction?

I1 = 13.0 A

P

3.0 m

I2 = 50.0 A



Tuesday, March 1, 2011

Loops of Wire / Solenoids



In the 4th Grade

s You learned that

coils with current

in them makemagnetic fields.

s The iron nail was

not necessary tocause the field; it

merely intensified

it.

I



Solenoid

s A solenoid is a coil of wire.s When current runs through the wire, it

causes the coil to become an“electromagnet”.

s Air-core solenoids have nothing insideof them.

s Iron-core solenoids are filled with iron tointensify the magnetic field.



Right Hand Rule for magnetic fields

around curved wires

1. Curve your fingers.

2. Place them along wireloop so that your fingerspoint in direction of current.

3. Your thumb gives thedirection of the magneticfield in the center of theloop, where it is straight.

4. Field lines curve around

and make completeloo s.

I

B



Magnetic Field around Curved

Current

B

Sample Problem



Sample ProblemWhat is the direction of the magnetic field

produced by the current I at A? At B?

I

AB



Magnetic Flux



Magnetic Flux

s The product of magnetic field and area.s Can be thought of as a total magnetic

“effect” on a coil of wire of a given area.

BA



Maximum Flux

s The area is aligned so that aperpendicular to the area points parallel

to the field

B

A



Minimum Flux

s The area is aligned so that aperpendicular to the area points

perpendicular to the field

B

A



Intermediate Flux

s The area is neither perpendicular nor isit parallel

B

A



Magnetic Flux

s ΦB = B A cosθ

– ΦB: magnetic flux in Webers (Teslameters2)

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

magnetic field.

s ΦB = B⋅ A

Sample Problem



Sample Problem

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 is4.2 mT.

a) Assume that the magnetic field is perpendicular to the areavector.

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

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

Sample Problem



Sample Problems Assume the angle is 40o, the magnetic field is 50 mT, and

the flux is 250 mWb. What is the radius of the loop?

B

A



Wednesday, March 2, 2011

Faraday’s Law of Induction



Induced Electric Potential

s A system will respond so as to opposechanges in magnetic flux.

s A change in magnetic flux will be partiallyoffset by an induced magnetic field whenever possible.

s Changing the magnetic flux through a wireloop causes current to flow in the loop.

s This is because changing magnetic fluxinduces an electric potential.



Faraday’s Law of Induction

s ε = -N∆ΦB/∆t

– ε: induced potential (V)

– N: # loops

– ΦB: magnetic flux (Webers, Wb)

– t: time (s)



A closer look …

s ε = -∆ΦB/∆t

s ε = -∆(BAcosθ)/∆t

– To generate voltage

• Change B

• Change A

• Change θ

S l P bl



Sample Problem

s A coil of radius 0.5 m consisting of 1000 loops is

placed in a 500 mT magnetic field such that the flux ismaximum. The field then drops to zero in 10 ms. What

is the induced potential in the coil?

S l P bl



Sample Problems 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 11seconds, it has a value of 600 mT. At time t = 11seconds, the field stops increasing. What is theinduced potential

s

A) at t = 0.5 seconds?s B) at t = 3.0 seconds?s C) at t = 12 seconds?



Lenz’s Law

s The current will flow in a direction so as

to oppose the change in flux.

s Use in combination with hand rule topredict current direction.

Sample Problem



ps The magnetic field is increasing at a rate of

4.0 mT/s. What is the direction of the

current in the wire loop?

Sample Problem



ps The magnetic field is decreasing at a rate of

4.0 mT/s. The radius of the loop is 3.0 m,

and the resistance is 4 Ω. What is themagnitude and direction of the current?



Monday, March 6, 2011

Motional EMF



Motional emf

s ε = BLv – B: magnetic field (T)

– L: length of bar moving through field – v: speed of bar moving through field.

s Bar must be “cutting through” field lines.It cannot be moving parallel to the field.

s This formula is easily derivable fromFaraday’s Law of Induction



Motional emf - derivation

s ε = ∆ΦB/∆t

s ε = ∆(BA) /∆t (assume cosθ = 1)

s ε = ∆(BLx) /∆t

s ε = BL∆x /∆t

s ε = BLv

Sample Problem



Sample Problems How much current flows through the resistor?

How much power is dissipated by the

resistor?

× × × × × × × × × × ×

× × × × × × × × × × ×

× × × × × × × × × × ×

× × × × × × × × × × ×

× × × × × × × × × × ×

× × × × × × × × × × ×

B = 0.15 T

v = 2 m/s

50 cm

3 Ω

Sample Problem



Sample Problems In which direction is the induced current

through the resistor (up or down)?

× × × × × × × × × × ×

× × × × × × × × × × ×

× × × × × × × × × × ×

× × × × × × × × × × ×

× × × × × × × × × × ×

× × × × × × × × × × ×

B = 0.15 T

v = 2 m/s

50 cm

3 Ω

Sample Problem



Sample Problems Assume the rod is being pulled so that it is traveling at

a constant 2 m/s. How much force must be applied to

keep it moving at this constant speed?

× × × × × × × × × × ×

× × × × × × × × × × ×

× × × × × × × × × × ×

× × × × × × × × × × ×

× × × × × × × × × × ×

B = 0.15 T

v = 2 m/s

50 cm

3 Ω

b magnetism power point

Documents