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General Physics (PHY 2140). Lecture 14. Electricity and Magnetism Magnetism Ampere’s law Applications of magnetic forces . http://www.physics.wayne.edu/~apetrov/PHY2140/. Chapter 19. Lightning Review. Last lecture: Magnetism Galvanometer Torque on a current loop.

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slide1

General Physics (PHY 2140)

Lecture 14

  • Electricity and Magnetism
      • Magnetism
          • Ampere’s law
          • Applications of magnetic forces

http://www.physics.wayne.edu/~apetrov/PHY2140/

Chapter 19

lightning review
Lightning Review
  • Last lecture:
  • Magnetism
    • Galvanometer
    • Torque on a current loop

Review Problem: A rectangular loop is placed in a uniform magnetic field with the plane of the loop perpendicular to the direction of the field. If a current is made to flow through the loop in the sense shown by the arrows, the field exerts on the loop:

1. a net force.

2. a net torque.

3. a net force and a net torque.

4. neither a net force nor a net torque.

19 7 motion of charged particle in magnetic field

q

v

F

19.7 Motion of Charged Particle in magnetic field

Bin

  • Consider positively charge particle moving in a uniform magnetic field.
  • Suppose the initial velocity of the particle is perpendicular to the direction of the field.
  • Then a magnetic force will be exerted on the particle…

´ ´´ ´ ´ ´ ´

´ ´ ´ ´ ´ ´ ´

´ ´ ´ ´ ´ ´ ´

´ ´ ´ ´ ´ ´ ´

r

Where is it directed?

… and make it follow a circular path.

Remember that

slide4

The magnetic force produces a centripetal acceleration.

The particle travels on a circular trajectory with a radius:

example 1 proton moving in uniform magnetic field
Example 1 : Proton moving in uniform magnetic field

A proton is moving in a circular orbit of radius 14 cm in a uniform magnetic field of magnitude 0.35 T, directed perpendicular to the velocity of the proton. Find the orbital speed of the proton.

Given:

r = 0.14 m

B = 0.35 T

m = 1.67x10-27 kg

q = 1.6 x 10-19 C

Recall that the proton’s radius would be

Thus

Find:

v = ?

example 2
Example 2:

Consider the mass spectrometer. The electric field between the plates of the velocity selector is 950 V/m, and the magnetic fields in both the velocity selector and the deflection chamber have magnitudes of 0.930 T. Calculate the radius of the path in the system for a singly charged ion with mass m=2.18×10-26 kg.

19 8 magnetic field of a long straight wire

I=0

19.8 Magnetic Field of a long straight wire
  • Danish scientist Hans Oersted (1777-1851) discovered (somewhat by accident) that an electric current in a wire deflects a nearby compass needle.
  • In 1820, he performed a simple experiment with many compasses that clearly showed the presence of a magnetic field around a wire carrying a current.

I

magnetic field due to currents
Magnetic Field due to Currents
  • The passage of a steady current in a wire produces a magnetic field around the wire.
    • Field form concentric lines around the wire
    • Direction of the field given by the right hand rule.
      • If the wire is grasped in the right hand with the thumb in the direction of the current, the fingers will curl in the direction of the field (second right-hand rule).
    • Magnitude of the field

I

slide9

Magnitude of the field

I

r

B

mocalled the permeability of free space

slide10

Ampere’s Law

Consider a circular path surrounding a current, divided in segments Dl, Ampere showed that the sum of the products of the field by the length of the segment is equal to mo times the current.

Andre-Marie Ampere

I

r

B

Dl

slide13

l

1

B2

I1

F1

2

d

I2

Force per unit length

definition of the si unit ampere
Definition of the SI unit Ampere

If two long, parallel wires 1 m apart carry the same current, and the magnetic force per unit length on each wire is 2x10-7 N/m, then the current is defined to be 1 A.

Used to define the SI unit of current called Ampere.

example 1 levitating a wire
Example 1: Levitating a wire

Two wires, each having a weight per units length of 1.0x10-4 N/m, are strung parallel to one another above the surface of the Earth, one directly above the other. The wires are aligned north-south. When their distance of separation is 0.10 mm what must be the current in each in order for the lower wire to levitate the upper wire. (Assume the two wires carry the same current).

l

1

I1

2

d

I2

slide16

Weight of wire per unit length:

mg/l = 1.0x10-4 N/m

Wire separation:

d=0.1 m

I1 = I2

Two wires, each having a weight per units length of 1.0x10-4 N/m, are strung parallel to one another above the surface of the Earth, one directly above the other. The wires are aligned north-south. When their distance of separation is 0.10 mm what must be the current in each in order for the lower wire to levitate the upper wire. (Assume the two wires carry the same current).

F1

1

I1

B2

mg/l

2

d

I2

l

example 2 magnetic field between the wires
Example 2: magnetic field between the wires

The two wires in the figure below carry currents of 3.00A and 5.00A in the direction indicated. Find the direction and magnitude of the magnetic field at a point midway between the wires.

5.00 A

3.00 A

20.0 cm

19 10 magnetic field of a current loop
19.10 Magnetic Field of a current loop
  • Magnetic field produced by a wire can be enhanced by having the wire in a loop.

Dx1

I

B

Dx2

19 11 magnetic field of a solenoid
19.11 Magnetic Field of a solenoid
  • Solenoid magnet consists of a wire coil with multiple loops.
  • It is often called an electromagnet.
solenoid magnet
Solenoid Magnet
  • Field lines inside a solenoid magnet are parallel, uniformly spaced and close together.
  • The field inside is uniform and strong.
  • The field outside is non uniform and much weaker.
  • One end of the solenoid acts as a north pole, the other as a south pole.
  • For a long and tightly looped solenoid, the field inside has a value:
solenoid magnet21
Solenoid Magnet

n = N/l : number of (loop) turns per unit length.

I : current in the solenoid.

example magnetic field inside a solenoid
Example: Magnetic Field inside a Solenoid.

Consider a solenoid consisting of 100 turns of wire and length of 10.0 cm. Find the magnetic field inside when it carries a current of 0.500 A.

N = 100

l = 0.100 m

I = 0.500 A

slide23

Comparison:Electric Field vs. Magnetic Field

Electric Magnetic

Source Charges Moving Charges

Acts on Charges Moving Charges

Force F = Eq F = q v B sin(q)

Direction Parallel E Perpendicular to v,B

Field Lines

Opposites Charges Attract Currents Repel