MAGNETISM – PART 3

CALENDAR

Monday – Recorded Lecture. Today – Brief review of the material + a few

problems. Today, Friday we will continue with chapter

20 materials. There will be a Quiz on Magnetism on Friday.

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Magnetism

ABOUT THOSE EXAMS - Grades look pretty bad. I will review on Friday

after I have a look at the papers. Each exam (both sections) had similar problems.

A Kirchoff Law Problem – simple A combine either capacitors or resistors and calculate

what was happening at one of them. A problem involving polarization – a thinker. A question on how much energy was required to bring

three charges together.

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Magnetism

THESE WERE THE HINTS I GAVE YOU! Anything in the three chapters is fair game. Read the sections on charge and charge effects very carefully.

We didn’t cover some of this in class. (Problem 2) Know the difference between Potential and Potential energy. Know how much work it takes to create a charge distribution.

We did it in class. (Problem 1) Know how to add capacitors and resistors and how to solve

simple circuit problems. (Problem 3) There WILL be a Kirchhoff's Law problem. (Problem 4) Coulomb’s Law and the addition of forces. Calculation of the

potential (scalar) Be sure to understand all of the HW problems that were

assigned – or not assigned!

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The force is perpendicularto the direction of motion.

The force has a constant magnitude = Bqv

This will produce circularmotion as in PHY2053.

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Magnetism

LOOK AT THE DIRECTION OF THE FORCE AND THE VELOCITY M

agnetism

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frequencyangular

cyclotron thecalled is This

v/ror r v:Recall

r

mv qvB

qvB is force magnetic Ther

mv Force lCentripeta

r

v on Accelerati lCentripeta

RECALL

2

2

2

m

Bq

r

v

Bq

mvr

Magnetism

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Bq

mvr

m

Bq

fperiodT

f

1

2

OFF ANGLE Magnetism

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TvP

PITCH

parallel

P

PROBLEM

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An electron at point A in the figure has a speed v0 of 1.4 x 106 m/s. Find

(a)the magnitude and direction of the magnetic field that will cause the electron to follow the semicircular path from A to B and

(b) the time required for the electron to move from A to B.

(c)What magnetic field would be needed if the particle were a proton instead of an electron?

m=9.1E-31 Kge=1.6E-19 C

Magnetism

FORCE ON A WIRE CARRYING A CURRENT IN A B FIELD

BilF

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A straight vertical wire carries a current of 1.20 A downward in a region between the poles of a large electromagnet where the field strength is 0.588 T and is horizontal. What are the magnitude and direction of the magnetic force on a 1.00 cm section of this wire if the magnetic-field direction is

(a)toward the east, (b) toward the south

Magnetism

• Novel applications have been devised to make use of the force that a magnetic field exerts on a conductor carrying current.

Magnetism

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CURRENT LOOP

Loop will tend to rotate due to the torque the field applies to the loop.

What is forceon the ends??

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THE LOOP (FROM THE TOP)

pivot

OBSERVATION

Force on Side 2 is outof the paper and that onthe opposite side is into the paper. No net forcetending to rotate the loopdue to either of these forces.

The net force on the loop is also zero,

Magnetism 17

THE OTHER SIDES1=F1 (b/2)Sin() =(B i a) x (b/2)Sin()

total torque on the loop is: 21

Total torque:

=(iaB) bSin() =iABSin()

(A=Area)

APPLICATION: THE MOTOR

If the conductor is a loop, the torque can create an electric motor.

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A circular coil of wire 8.6 cm in diameter has 15 turns and carries a current of 2.7 A. The coil is in a region where the magnetic field is 0.56 T.

What orientation of the coil gives the maximum torque on the coil.

What is this maximum torque?

Magnetism

Magnetism 23

ANOTHER APPLICATIONTHE GALVANOMETER

CURRENTS CAUSE MAGNETIC FIELDS

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MAGNETIC FIELD OF LONG STRAIGHT CONDUCTOR –

• Placed over a compass, the wire would cause the compass needle to deflect. This was the classic demonstration done by Oersted as he demonstrated the effect.

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RESULT Magnetism

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r(exact) 104

2

70

0

A

Tm

r

IB

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FORCE BETWEEN TWO CURRENT CARRYING CONDUCTORS

First wire produces a magnetic field at the second wire position.

The second wire therefore feels a force = Bil

TWO WIRES Magnetism

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r

II

l

F

lIr

IlBIF

r

I

2

2

2B

First Wire From B

210

210

2

10

CURRENTS IN A LOOP – M

agnetism

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FIELD OF A CURRENT LOOP Magnetism

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R

NIB

20

N turns of wire

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SOLONOID Magnetism

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Length

Turns of

number Total0

L

Nn

nIB

B=~0 outside

THE SOLENOID – M

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r

NIB

20

B=0 outside

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