Chapter 20

Magnetism

Units of Chapter 20

• Magnets and Magnetic Fields

• Electric Currents Produce Magnetic Fields

• Force on an Electric Current in a Magnetic Field; Definition of B

• Force on Electric Charge Moving in a Magnetic Field

• Magnetic Field Due to a Long Straight Wire

• Force between Two Parallel Wires

Units of Chapter 20

• Solenoids and Electromagnets

• Ampère’s Law

• Torque on a Current Loop; Magnetic Moment

• Applications: Galvanometers, Motors, Loudspeakers

• Mass Spectrometer

• Ferromagnetism: Domains and Hysteresis

20.1 Magnets and Magnetic Fields

Magnets have two ends – poles – called north and south.

Like poles repel; unlike poles attract.

20.1 Magnets and Magnetic Fields

However, if you cut a magnet in half, you don’t get a north pole and a south pole – you get two smaller magnets.

20.1 Magnets and Magnetic Fields

Magnetic fields can be visualized using magnetic field lines, which are always closed loops.

20.1 Magnets and Magnetic Fields

The Earth’s magnetic field is similar to that of a bar magnet.

Note that the Earth’s “North Pole” is really a south magnetic pole, as the north ends of magnets are attracted to it.

20.1 Magnets and Magnetic Fields

A uniform magnetic field is constant in magnitude and direction.

The field between these two wide poles is nearly uniform.

20.2 Electric Currents Produce Magnetic Fields

Experiment shows that an electric current produces a magnetic field.

20.2 Electric Currents Produce Magnetic Fields

The direction of the field is given by a right-hand rule.

20.3 Force on an Electric Current in a Magnetic Field; Definition of B

A magnet exerts a force on a current-carrying wire. The direction of the force is given by a right-hand rule.

20.3 Force on an Electric Current in a Magnetic Field; Definition of B

The force on the wire depends on the current, the length of the wire, the magnetic field, and its orientation.

(20-1)

This equation defines the magnetic field B.

20.3 Force on an Electric Current in a Magnetic Field; Definition of B

Unit of B: the tesla, T.

1 T = 1 N/A·m.

Another unit sometimes used: the gauss (G).

1 G = 10-4 T.

Example 20-1A wire carrying a 30 A current has a length of 12 cm between the pole faces of a magnet at an angle of 60 degrees. The magnetic field is approximately uniform at 0.90 T. We ignore the field beyond the pole pieces. What is the magnitude of the force on the wire?

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l =12 cm, I = 30 A, B = 0.90 T, θ = 60o

F = IlBsinθ

F = (30 A)(0.12 m)(0.90 T)sin60

F = 2.8 N

Example 20-1A rectangular loop of wire hangs vertically as shown. A magnetic field B is directed horizontally, perpendicular to the wire, and points out of the page at all points. The magnetic field is very nearly uniform along the horizontal portion of wire ab (length = 10.0 cm) which is near the center of the gap of a large magnet producing the field. The top portion of the wire loop is free of the field. The loop hangs from a balance which measures a downward force (in addition to the gravitational force) of F=3.48x10-2 N when the wire carries a current of 0.245 A. What is the magnitude of the magnetic field B?

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F = IlBsinθ, θ = 90o

B =F

Il=

3.48x10-2 N

(0.245 A)(0.100 m)

B =1.42 T

20.4 Force on Electric Charge Moving in a Magnetic Field

The force on a moving charge is related to the force on a current:

(20-3)

Once again, the direction is given by a right-hand rule.

20.4 Force on Electric Charge Moving in a Magnetic Field

If a charged particle is moving perpendicular to a uniform magnetic field, its path will be a circle.

20.4 Force on Electric Charge Moving in a Magnetic Field

Problem solving: Magnetic fields – things to remember

1. The magnetic force is perpendicular to the magnetic field direction.

2. The right-hand rule is useful for determining directions.

3. Equations in this chapter give magnitudes only. The right-hand rule gives the direction.

20.4 Force on Electric Charge Moving in a Magnetic Field

Example 20-4A proton having a speed of 5.0x106 m/s in a magnetic field feels a force of 8.0x10-14 N toward the west when it moves vertically upward. When moving horizontally in a northern direction, it feels zero force. Determine the magnitude and direction of the magnetic field in this region. (The charge on a proton is q=+e=1.6x10-19 C.)

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F = qvBsinθ, θ = 90o

B =F

qv=

8.0x10-14 N

(1.6x10-19 C)(5.0x106 s)

B = 0.10 T

No force when moving north means the field must be in a north-south direction. In order for the force to pint west in (a), the field must point north (which is into the page in this figure) from the right hand rule.

Example 20-5An electron travels at 2.0x107 m/s in a plane perpendicular to a uniform 0.010 T magnetic field. Describe its path qualitatively.

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ΣF = ma

qvB = mv2

r

r =mv

qB=

(9.1x10-31 kg)(2.0x107 m/s)

(1.6x10-19 C)(0.010 T)

So the electron moves in a circle of radius

1.1x10-2 m, or 1.1 cm.

20.5 Magnetic Field Due to a Long Straight Wire

The field is inversely proportional to the distance from the wire:

(20-6)

The constant μ0 is called the permeability of free space, and has the value:

Example 20-7An electric wire in the wall of a building carries a dc current of 25 A vertically upward. What is the magnetic field due to this current at a point P 10 cm due north of the wire?

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B =μ0I

2πr

B =(4πx10-7 Tm/A)(25 A)

(2π )(0.10 m)

B = 5.0x10-5 T

Note: The wire’s field has about the same magnitude as Earth’s so a compass would not point north but in a northwesterly direction.

Note: Most electrical wiring in buildings consists of cables with two wires in each cable. Since the two wires carry current in opposite direction, their magnetic fields will cancel to a large extent.

Example 20-8Two parallel straight wires 10.0 cm apart carry currents in opposite directions. Current I1=5.0 A is out of the page, and I2=7.0 A is into the page. Determine the magnitude and direction of the magnetic field halfway between the two wires.

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From right - hand - rule 1, both fields point

upwards at the midpoint, which is 0.050 m

from each wire.

B1 =μ0I1

2πr=

(4πx10-7 Tm/A)(5.0 A)

(2π )(0.050 m)= 2.0x10-5 T

B2 =μ0I2

2πr=

(4πx10-7 Tm/A)(7.0 A)

(2π )(0.050 m)= 2.8x10-5 T

B = B1 +B2 = 4.8x10-5 T