Chapter 30 Ampere’s Law

1 / 15

# Chapter 30 Ampere’s Law - PowerPoint PPT Presentation

Chapter 30 Ampere’s Law. PHYS 2326-19. Concepts to Know. Magnetic Forces between wires Ampere’s law Solenoid Toroid. Magnetic Forces Between Wires. Chapter 30.2 Similar to forces between two charged particles

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### Chapter 30 Ampere’s Law

PHYS 2326-19

Concepts to Know
• Magnetic Forces between wires
• Ampere’s law
• Solenoid
• Toroid
Magnetic Forces Between Wires
• Chapter 30.2
• Similar to forces between two charged particles
• Give 2 parallel wires 1 & 2 separated by distance a with currents I1 & I2 What are the forces between them?

B2

1

I1

F1

2

I2

a

Magnetic Force Between 2 Wires
• Wire 2 generates a field B2
• For a long straight wire B2 is from eqn 30.5
• Force F1 on wire 1 comes from eqn 29.10
• Using magnitudes

eqn 30.12

Using right hand rules for directions

B2 is out of plane and F1 is down towards wire 2

Ampere’s Law
• The line integral of B·ds (B dot ds) is equal to μo I (the permeability of free space times the current I)
The Toroid
• A ring or torus wrapped with a wire
• Create amperian loop – dashed line
• by symmetry (assume wire uniform all the way around) – the field is constant B and tangent to it so B·ds = B ds
• Wire passes through N times (4 shown)
The Toroid
• Since there are N turns, total current through the loop is NI
• Apply Ampere’s law

.

.

.

.

.

.

.

x

x

x

x

x

x

The Solenoid
• Ideal solenoid is when length >> radius
• B is uniform and parallel inside ideal solenoid
• Consider loop 2 a rectangle w*l in area
• Side 3 is far away so B = 0, side 2&4 B is perpendicular with the sides
• Side 1 is parallel and in a uniform field B so that the magnitude is Bl

B

loop 2

2

3

l

1

4

w

loop 1

Example 1
• Given a rectangular loop of 10cm x 20cm (a by b) with counter clockwise current I=2A what is the magnitude and direction of magnetic field at the center point P
• B direction right hand rule = out of page
• B = Bab+Bbc+Bcd+Bda = 2Bab + 2Bbc

a

d

I

P

b

c

Example 1

From eqn 30.4 for the field produced by a long wire. The book shows sines which is opposite/adjacent

• B = 8.78E-6 T
Example 2
• Long straight vertical wire carries 10A

A rectangular coil of wire located near it carries 5 A, a = 0.10m, b=0.30m, c=0.50m. Force on loop?

Force on top = -Force on bottom

b

c

I2

I1

a

x

Example 3

Use Ampere’s Law to determine magnetic field

Choose a circle centered around the wire so B is tangent at every point and uniform in distance from the wire

B

dl

r

I

Example 3
• Dot product B·l = Bdl
• Same magnitude at each point (symmetry)
• B is constant so
Example 4Inside and outside a Toroid
• Find the magnetic field a) inside and b) outside a tightly wound toroid of N turns