phys 1112 ch 21 notes

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CHAPTER 21 MAGNETIC FORCES AND FIELDS

Magnetic forces and magnetic fields are associated with

moving electric charge.In the case of a permanent magnet made of iron, nickel,

cobalt or some combination of these and other elements,

the magnetic field comes from the alignment of electron

spin axes.

These elements are called ferromagnetic because they

can be magnetized. They have unpaired electrons that

can be aligned in regions called domains. When enough

of these domains within an object are aligned in thesame direction, the object becomes magnetic and has

north and south magnetic poles.

If the magnet is allowed to align itself with the earth's

magnetic field, the end that points to the earth's north

pole is called a north pole. The end that points south is

called a south pole.

This has caused some confusion.The law of poles says like poles repell and unlike poles

attract. This can be demonstrated with two magnets of

similar strength.

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Due to the way the poles are named, the earth's

magnetic south pole is near the earth's geographic northpole. In about 100,000 years, the earth's magnetic poles

will reverse and our compasses will all point in the

opposite direction.

The direction of a magnetic field is represented by

magnetic field lines. They are drawn pointing from

North pole to South pole.

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The relative distance between the field lines indicatesthe magnetic field strength. Where they are close

together, it is stronger. Where they are parallel, it is

constant.

When a charge moves in a magnetic field, it experiences

a magnetic force. If it is at rest relative to the magnetic

field, it does not. This happens because a moving charge

generates a magnetic field that interacts with any other

magnetic field that is present. A charge at rest does not

generate a magnetic field and is not repelled by or

attracted to a magnet.

Also, a moving charge must have a component of

motion perpendicular to the direction of the magnetic

field causing the force. The magnetic field produced bythe moving charge forms a series of concentric circles

around the charge.

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In the diagram below, the component of this magnetic

field that opposes the external field will cause an

attractive force to exist.(N attracts S) The componentoriented in the same direction will cause a repulsive

force to exist. In this diagram both of these forces result

in a net force directed upward. This force is always

perpendicular to the direction of the external magnetic

field and the velocity vector of the particle.

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If the velocity vector is at some angle other than 90

with respect to the magnetic field direction, we must

find the perpendicular component of the velocity by

using Vsin where is the angle between the velocityvector and the direction of the magnetic field.

The equation for this relationship is:

B = F/(q0Vsin)

where B is the magnetic field strength in Teslas, F is the force

acting on the charged particle in Newtons, q0 is the charge, and is the angle between the velocity vector and B.

One Tesla is one Newton per Coulomb meter per second. It is

usually described as one Newton per ampere meter.

The Tesla is a large unit of magnetic field strength and the

smaller unit, the Gauss is often used. The Gauss is

approximately equal to the strength of the earth's magnetic

field near the surface of the earth. 1 G = 10-4 T.

Example

A particle with a charge of 8.4 C and a speed of 45 m/s enters

a uniform magnetic field whose magnitude is 0.30 T. For each

of the cases in the drawing, find the magnitude and direction of

the magnetic force acting on the particle.

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Motion of a Charged Particle in a Magnetic Field

Motion o f a Charge in Magnetic and Electric Fields

An important aspect of motion of a charged particle in a

magnetic field involves a comparison to motion in an electric

field.

In both cases, a positively charged particle enters a uniformfield with a velocity at 90 to the direction of the field.

In the case of the electric field the particle turns in the

direction of the field. Since the force exerted on the particle is

in the same direction as a component of the particle's velocity,
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work is done on the particle, it gains kinetic energy and speeds

up.

In the case of the magnetic field, the force causes the particle to

turn in a direction that forms a 90 angle with both the velocity

and the magnetic field. Since there is no component of velocityin the same direction as the force, no work is done and the

kinetic energy, therefore the speed remains constant.

When a charged particle enters a uniform magnetic field so

that its velocity is oriented at 90 to the direction of the field,

its path will become a circle.

The magnetic force will always remain perpendicular to the

velocity and is directed toward the center of the circle.

The radius of the circular path can be found using the equation

for centripetal force.

F = mv2/r

F = qBvsin

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mv2/r = qBvsin

Since v is at right angles to B, = 90 and sin = 1.

mv = qBr

r = mv/qB

which gives us the equation for the radius of the path.

Example

A charged particle enters a uniform magnetic field and follows

the circular path shown in the drawing. Is the charge on the

particle positive or negative? If the particle's speed is 140 m/s,

the magnitude of the magnetic field is 0.48 T, the radius of the

path is 960 m, and the charge on the particle is 820 C, find the

mass of the particle.

The mass spectrometer applies this relationship to separate

charged particles with different masses. Instead of the velocity

of the particle, the potential difference V which is applied

between the ion source and metal plate is known.

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The ions pass through the metal plate, acquiring kinetic energy

equal to Vq as they enter a region with a constant magnetic

field, B where they follow a semicircular path to the detector.

Since Vq = mv2, we can replace v in the equation mv = qBr,

and derive the equation for the mass spectrometer which is:

m = (er2/2V)B2

which shows that the mass of an ion reaching the detector at 2ris directly proportional to B2. Varying the field strength allows

ions of different masses to be detected.

Example

The ion source in a mass spectrometer produces singly and

doubly ionized species. The difference in mass between thesespecies is too small to be detected. Both species experience the

same potential difference and magnetic field. Find the ratio of

the radius of the path of particle 1(+e) to that of particle

2(+2e).

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The Force on a Current in a Magnetic Field

When a charge moves through a magnetic field, the force it

experiences is proportional to the magnitude of the charge, its

speed, the magnetic field strength, and the sine of the anglebetween the direction of the magnetic field and the direction of

the velocity of the particle.

The force on an electric current in a wire can be calculated

much the same way since a current consists of charges moving

through a wire.

F = qvBsin

F = (q/t)(vt)sin

Since q/t = I and vt = L(length of the wire), the equation

becomes:

F = ILBsin

Example

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A square coil of wire containing a single turn is placed in a

uniform 0.25 T magnetic field. Each side has a length of 0.32 m

and the current in the coil is 12 A. Determine the magnitude of

the magnetic force on each of the four sides.

The Torque on a Current Carrying Coil

In the previos problem we calculated the force acting on the

four sides of a square coil of wire carrying a current in a

magnetic field. The coil was oriented so that the direction of

the magnetic field was at right angles to the direction of theplane of the coil.

If we attach this loop to an axis with a set of bearings in the

same plane as the loop and through its center the forces

exerted on the loop will make it rotate until the normal to the

plane of the loop aligns with the external magnetic field.

If we multiply the force generated by the radius of the loop, we

get the torque generated by one force. Since this arrangement

acts like a couple, the torque is equal to the force generatedmultiplied by the entiredistance across the coil.

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= Fw

= ILBwsin

since F = ILB and = the angle between the force and the lever

arm connecting the axis of rotation to the wire in the coil. Also,

Lw = area of the coil, so

= IABsin

and if there are N coils instead of 1,

= NIABsin

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The torque depends on the geometry of the coil, the current

through it, and the strength of the external magnetic field

causing the rotation.

NIA is called the magnetic moment of the coil and can be usedto compare torque generated by different coils placed in the

same magnetic field.

These equations apply to circular as well as square or

rectangular coils.

Example

The maximum torque experienced by a coil in a 0.75 Tmagnetic field is 8.4 x 10-4 Nm. The coil is circular and consists

of 1 turn. If the current in the coil is 3.7 A, find the length of

wire from which the coil is made.

Electric motors work because of this torque generated by the

interaction of the magnetic field associated with the current in

a coil on the armature with the external magnetic field

produced by the field magnets.

Magnetic Fields Produced by Currents

Magnetic Fiel d Produced by Long Straight Current-Carrying

Wire

Any time an electric current flows in a wire, it produces a

magnetic field.

In a long, straight wire, the direction of the magnetic field can

be determined using the right hand rule.
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Point the thumb of your right hand in the direction of the

current in the wire and curl your fingers. Your fingers now

point in the direction of the magnetic field caused by thecurrent in the wire.

The strength of the field is directly proportional to the current

in the wire and inversely proportional to the distance from the

wire. The equation is:

B = 0I/2r

where 0 is a proportionality constant called thepermeability of free space. Its value is 4 x 10-7 Tm/A.

Example

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A long,straight wire carries a current of 48 amps. The strength

of the magnetic field produced by this current at a certain

point is 8.0 x 10-5 T. Find the shortest distance from the point

to the wire.

When two current carrying wires are near each other, each

one exerts a force on the other. This can be a repulsive force or

an attractive force depending on the current direction.

Consider wire 1 to be fixed. In diagram (a) wire 2 experiences

a force due to the magnetic field generated by wire 1 and the

current flowing through wire 2. Right hand rule # 2, when

applied to wire 1, tells us that the magnetic field at wire 2 acts

upward.

Right hand rule # 1 when applied to wire 2 tells us that the

force acting on wire 2 is to the right, perpendicular to both the

current direction and the magnetic field direction.

The magnitude of the force between two current carrying

wires can be calculated using the equations:

F2 = I2LBsin

Since this is the formula for the force acting on a wire

with a current of I2 of length L in a magnetic field of

strength B.

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B = 0I1/2r

This equation is used to calculate the magnetic field

strength due to current I1 in the first wire. If wesubstitute into the first equation for B we get:

F = I2L (0I1/2r)sin

F = I1I2L(0/2r)

where I1 and I2 are the currents through each wire, L is the

length of each wire, and r is the perpendicular distancebetween them.

If the two currents are in the same direction, the force is

attractive. If they are in opposite directions, it is repulsive.

Example

Two rigid rods are parallel to each other and the ground. Theycarry the same current in the same direction. The length of

each rod is 0.85 meters and the mass of each is 0.073 Kg. One

rod floats beneath the other rod at a distance of 0.0082 meters.

Find the current in each rod.

Magnetic Field Associated with a Loop of Wire

Magnetic Field Produced by a Current Loop

When a wire is bent into a circular loop, the magnetic field

lines generated resemble those of a bar magnet.

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The direction of the magnetic field at the center of a loop canbe determined using right hand rule 2. The magnitude of the

magnetic field can be determined with the equation:

B = N0I/2R

Example

Find the radius of a circular single loop of wire so that themagnetic field at the center is 1.8 x 10-8 T when the loop carries

a current of 12 A.

Magnetic Field Produced by a Solenoid

A solenoid is a long coil of wire shaped like a helix

approximating a series of circles. The magnetic field inside asolenoid is constant and can be calculated using the formula:

B = 0nI

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where n is the number of turns per meter of length and I is the

current. If the length of the solenoid is much greater than its

diameter the magnetic field outside the solenoid is nearly zero.

Solenoids are useful as electromagnets, electrically controlledswitches, in CRT image production and in magnetic resonance

imaging.

This MRI angiogram shows bleeding into the lower central

part of the brain.

Ampere's Law

For any current geometry that produces a magnetic field that

does not change in time,

BparL = 0I

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This means that the sum of all the products of small distances

an the component of the magnetic field parallel to them is

equal to the product of the permeability of free space and the

current through the surface bounded by that total closed path.

P 678 Questions 2, 4, 5, 6, 7, 9, 10, 14, 15, 19

P 679 Problems 1, 3, 5, 6, 11, 12, 17, 18, 27, 29, 31, 36, 37, 39,

46, 47, 49, 53, 67, 71

phys 1112 ch 21 notes

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