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## PowerPoint Slideshow about ' Magnetism – Part 3' - ferdinand-reynolds

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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.

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.

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 circuitproblems. (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!

to the direction of motion.

The force has a constant

magnitude = Bqv

This will produce circular

motion as in PHY2053.

PROBLEM

- An electron at point A in the figure has a speed v0 of 1.4 x 106 m/s. Find
- the magnitude and direction of the magnetic field that will cause the electron to follow the semicircular path from A to B and
- the time required for the electron to move from A to B.
- What magnetic field would be needed if the particle were a proton instead of an electron?

m=9.1E-31 Kg

e=1.6E-19 C

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

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

Current Loop force that a magnetic field exerts on a conductor carrying current.

What is force

on the ends??

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

pivot force that a magnetic field exerts on a conductor carrying current.

The Loop (From the top)OBSERVATION

Force on Side 2 is out

of the paper and that on

the opposite side is into

the paper. No net force

tending to rotate the loop

due to either of these forces.

The net force on the loop is

also zero,

The other sides force that a magnetic field exerts on a conductor carrying current.

t1=F1 (b/2)Sin(q)

=(B i a) x (b/2)Sin(q)

total torque on

the loop is: 2t1

Total torque:

t=(iaB) bSin(q)

=iABSin(q)

(A=Area)

Application: The motor force that a magnetic field exerts on a conductor carrying current.

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

A circular coil of wire 8.6 cm in diameter has force that a magnetic field exerts on a conductor carrying current.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?

Another force that a magnetic field exerts on a conductor carrying current.ApplicationThe Galvanometer

Currents Cause Magnetic Fields force that a magnetic field exerts on a conductor carrying current.

Magnetic field of long straight conductor – force that a magnetic field exerts on a conductor carrying current.

- 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.

Result force that a magnetic field exerts on a conductor carrying current.

r

Force Between Two Current Carrying Conductors force that a magnetic field exerts on a conductor carrying current.

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

The second wire therefore feels a force = Bil

Two Wires force that a magnetic field exerts on a conductor carrying current.

Currents in a loop – force that a magnetic field exerts on a conductor carrying current.

Field of a Current Loop force that a magnetic field exerts on a conductor carrying current.

N turns of wire

Solonoid force that a magnetic field exerts on a conductor carrying current.

B=~0 outside

The solenoid – force that a magnetic field exerts on a conductor carrying current.

B=0 outside

force that a magnetic field exerts on a conductor carrying current.

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