Physics 201Professor P. Q. Hung

311B, Physics Building

Physics 201 – p. 1/30

Magnetic Fields

A proton has a speed of 106 m/s and enters aregion with a uniform magnetic field. It makes anangle of 450 with respect to the direction of themagnetic field ~B. Find its subsequent motion.

What is a magnetic field?

Subsequent motion? Is there any force? Areelectrically neutral particles affected by amagnetic field? Or only charged particles areaffected?

Physics 201 – p. 2/30

Magnetic Fields

A proton has a speed of 106 m/s and enters aregion with a uniform magnetic field. It makes anangle of 450 with respect to the direction of themagnetic field ~B. Find its subsequent motion.

What is a magnetic field?

Subsequent motion? Is there any force? Areelectrically neutral particles affected by amagnetic field? Or only charged particles areaffected?

Physics 201 – p. 2/30

Magnetic Fields

Some facts

We have seen how magnets influence themotion of particles which are charged, howthey interact with metallic objects, etc...

A magnet has a north pole and a south polewhich do not actually coincide with thegeographical poles (from the axis of rotation).⇒Like poles repel and unlike poles attracteach other.

Physics 201 – p. 3/30

Magnetic Fields

Some facts

We have seen how magnets influence themotion of particles which are charged, howthey interact with metallic objects, etc...

A magnet has a north pole and a south polewhich do not actually coincide with thegeographical poles (from the axis of rotation).⇒Like poles repel and unlike poles attracteach other.

Physics 201 – p. 3/30

Magnetic Fields

Some facts

Just like the case of electrostatics, there is amagnetic field surrounding a magnet.

One can draw magnetic field lines startingfrom the “North” pole and ending in the“South” pole.

We will denote a magnetic field by ~B.

Physics 201 – p. 4/30

Magnetic Fields

Some facts

Just like the case of electrostatics, there is amagnetic field surrounding a magnet.

One can draw magnetic field lines startingfrom the “North” pole and ending in the“South” pole.

We will denote a magnetic field by ~B.

Physics 201 – p. 4/30

Magnetic Fields

Some facts

Just like the case of electrostatics, there is amagnetic field surrounding a magnet.

One can draw magnetic field lines startingfrom the “North” pole and ending in the“South” pole.

We will denote a magnetic field by ~B.

Physics 201 – p. 4/30

Magnetic Fields

Magnetic force between two magnets

Physics 201 – p. 5/30

Magnetic Fields

Always two poles, North and South

Physics 201 – p. 6/30

Magnetic Fields

Magnetic field lines of a bar magnet

Physics 201 – p. 7/30

Magnetic Fields

Refrigerator magnet

Physics 201 – p. 8/30

Magnetic Fields

Magnetic field lines of the Earth

Physics 201 – p. 9/30

Magnetic Fields

A proton has a speed of 106 m/s and enters aregion with a uniform magnetic field. It makes anangle of 450 with respect to the direction of themagnetic field ~B. Find its subsequent motion.

What if the proton enters the region with auniform magnetic field with zero speed?Nothing happens!

What if it enters that region with a velocitywhich is either parallel to or antiparallel to themagnetic field? It will experience no force.

Physics 201 – p. 10/30

Magnetic Fields

A proton has a speed of 106 m/s and enters aregion with a uniform magnetic field. It makes anangle of 450 with respect to the direction of themagnetic field ~B. Find its subsequent motion.

What if the proton enters the region with auniform magnetic field with zero speed?Nothing happens!

What if it enters that region with a velocitywhich is either parallel to or antiparallel to themagnetic field? It will experience no force.

Physics 201 – p. 10/30

Magnetic Fields

When neither of the previous two casesapplies, it is found that the force is directlyproportional to the magnitudes of themagnetic field, the charge, and the speed.

It is found that this force is maximal when thevelocity of the proton is perpendicular to thedirection of the magnetic field.

Physics 201 – p. 11/30

Magnetic Fields

When neither of the previous two casesapplies, it is found that the force is directlyproportional to the magnitudes of themagnetic field, the charge, and the speed.

It is found that this force is maximal when thevelocity of the proton is perpendicular to thedirection of the magnetic field.

Physics 201 – p. 11/30

Magnetic Fields

Force on moving charge

~F = q~v × ~B. zero when ~v = 0 and/or when~v ‖ ~B

Magnitude: F = qv B sin θ. θ: Angle between~v and ~B

F is maximal when θ = 900.

B = Fq v sin θ

. Unit: 1 tesla (T) = 1 N/(A.m).

1 gauss = 10−4T

Physics 201 – p. 12/30

Magnetic Fields

Force on moving charge

~F = q~v × ~B. zero when ~v = 0 and/or when~v ‖ ~B

Magnitude: F = qv B sin θ. θ: Angle between~v and ~B

F is maximal when θ = 900.

B = Fq v sin θ

. Unit: 1 tesla (T) = 1 N/(A.m).

1 gauss = 10−4T

Physics 201 – p. 12/30

Magnetic Fields

Force on moving charge

~F = q~v × ~B. zero when ~v = 0 and/or when~v ‖ ~B

Magnitude: F = qv B sin θ. θ: Angle between~v and ~B

F is maximal when θ = 900.

B = Fq v sin θ

. Unit: 1 tesla (T) = 1 N/(A.m).

1 gauss = 10−4T

Physics 201 – p. 12/30

Magnetic Fields

Force on moving charge

~F = q~v × ~B. zero when ~v = 0 and/or when~v ‖ ~B

Magnitude: F = qv B sin θ. θ: Angle between~v and ~B

F is maximal when θ = 900.

B = Fq v sin θ

. Unit: 1 tesla (T) = 1 N/(A.m).

1 gauss = 10−4T

Physics 201 – p. 12/30

Magnetic Fields

Force on a moving charge

Physics 201 – p. 13/30

Magnetic Fields

Force on a moving charge

Physics 201 – p. 14/30

Magnetic Fields

A proton has a speed of 106 m/s and enters aregion with a uniform magnetic field. It makes anangle of 450 with respect to the direction of themagnetic field ~B. Find its subsequent motion.

Subsequent motion? Use Newton’s law:a = F

mp

= qvB sin θ

mp

For e.g. B = 0.5T :

a =(1.6×10−19C)(106 m

s)(0.5T )(

√

2

2)

1.67×10−27 kg= 3.4 × 1013m/s2

Physics 201 – p. 15/30

Magnetic Fields

A proton has a speed of 106 m/s and enters aregion with a uniform magnetic field. It makes anangle of 450 with respect to the direction of themagnetic field ~B. Find its subsequent motion.

Subsequent motion? Use Newton’s law:a = F

mp

= qvB sin θ

mp

For e.g. B = 0.5T :

a =(1.6×10−19C)(106 m

s)(0.5T )(

√

2

2)

1.67×10−27 kg= 3.4 × 1013m/s2

Physics 201 – p. 15/30

Magnetic Fields

What happens to the acceleration if it were anelectron instead?

a =(1.6×10−19C)(106 m

s)(0.5T )(

√

2

2)

9.11×10−31 kg= 0.6 × 1017m/s2.

Easier to accelerate electrons than protonsbecause they are lighter.

Physics 201 – p. 16/30

Magnetic Fields

What happens to the acceleration if it were anelectron instead?

a =(1.6×10−19C)(106 m

s)(0.5T )(

√

2

2)

9.11×10−31 kg= 0.6 × 1017m/s2.

Easier to accelerate electrons than protonsbecause they are lighter.

Physics 201 – p. 16/30

Magnetic Fields

How does the motion of the proton look like?

Since the velocity of the proton makes anangle of 450 with respect to the direction ofthe magnetic field ~B, it has a componentv cos θ along the direction of ~B. Thiscomponent is unchanged.

The perpendicular component ~vperpendicular ischanged in direction due to the above force⇒ Circular motion around ~B. Combined withthe unchanged parallel component, themotion looks like a helix!

Physics 201 – p. 17/30

Magnetic Fields

How does the motion of the proton look like?

Since the velocity of the proton makes anangle of 450 with respect to the direction ofthe magnetic field ~B, it has a componentv cos θ along the direction of ~B. Thiscomponent is unchanged.

The perpendicular component ~vperpendicular ischanged in direction due to the above force⇒ Circular motion around ~B. Combined withthe unchanged parallel component, themotion looks like a helix!

Physics 201 – p. 17/30

Magnetic Fields

How does the motion of the proton look like?

Physics 201 – p. 18/30

Magnetic Fields

What if the proton comes in with a velocity whichis perpendicular to the magnetic field?

In this case, there is no parallel component ofthe velocity ⇒ circular motion in a planeperpendicular to the magnetic field. ⇒F = qvB = mv2

r= mω2r ⇒

r = mvqB

= PqB

: Radius of circular motion

ω = vr

= qB

mCyclotron angular frequency

T = 2πω

= 2πmqB

Physics 201 – p. 19/30

Magnetic Fields

What if the proton comes in with a velocity whichis perpendicular to the magnetic field?

In this case, there is no parallel component ofthe velocity ⇒ circular motion in a planeperpendicular to the magnetic field. ⇒F = qvB = mv2

r= mω2r ⇒

r = mvqB

= PqB

: Radius of circular motion

ω = vr

= qB

mCyclotron angular frequency

T = 2πω

= 2πmqB

Physics 201 – p. 19/30

Magnetic Fields

What if the proton comes in with a velocity whichis perpendicular to the magnetic field?

In this case, there is no parallel component ofthe velocity ⇒ circular motion in a planeperpendicular to the magnetic field. ⇒F = qvB = mv2

r= mω2r ⇒

r = mvqB

= PqB

: Radius of circular motion

ω = vr

= qB

mCyclotron angular frequency

T = 2πω

= 2πmqB

Physics 201 – p. 19/30

Magnetic Fields

What if the proton comes in with a velocity whichis perpendicular to the magnetic field?

In this case, there is no parallel component ofthe velocity ⇒ circular motion in a planeperpendicular to the magnetic field. ⇒F = qvB = mv2

r= mω2r ⇒

r = mvqB

= PqB

: Radius of circular motion

ω = vr

= qB

mCyclotron angular frequency

T = 2πω

= 2πmqB

Physics 201 – p. 19/30

Magnetic Fields

Example of circular motionAn electron and a proton move in circular orbitsin a plane perpendicular to a uniform magneticfield ~B. Find the ratio of the radii of their circularorbits when the electron and the proton have (a)the same momentum and (b) the same kineticenergy.

(a) r = mvqB

= PqB

⇒ re

rp

= Pe

Pp

= 1 because qB is

the same for both.

Physics 201 – p. 20/30

Magnetic Fields

Example of circular motion

(b) K = P 2

2m ⇒ P =√

2mK ⇒ re

rp

= Pe

Pp

=√

2meKe

2mpKp

=√

me

mp

=√

9.11×10−31 kg1.67×10−27 kg

= 2.3 × 10−2

Physics 201 – p. 21/30

Magnetic Fields

Some applications: Velocity selector

In some situations, it is important for thecharged particles to all move with the samevelocity. Let them pass through a region inwhich there is a uniform magnetic field (saypointing into the plane of the paper) and auniform electric field perpendicular to it (saypointing downward).

For q > 0, q~v × ~B points upwardand q ~E points downward.

Physics 201 – p. 22/30

Magnetic Fields

Some applications: Velocity selector

In some situations, it is important for thecharged particles to all move with the samevelocity. Let them pass through a region inwhich there is a uniform magnetic field (saypointing into the plane of the paper) and auniform electric field perpendicular to it (saypointing downward).

For q > 0, q~v × ~B points upwardand q ~E points downward.

Physics 201 – p. 22/30

Magnetic Fields

Some applications: Velocity selector

~E and ~B can be chosen so that these twoforces cancel ⇒ qvB = qE⇒ v = E

B

Physics 201 – p. 23/30

Magnetic Fields

Some applications: Electromagnetic flowmeteras a Velocity selector

Physics 201 – p. 24/30

Magnetic Fields

Some applications: Mass spectrometer

After passing through a velocity selector, theparticles go through a second region withuniform magnetic field B0 in the samedirection as the one in the velocity selector.They bend around and using the expressionfor the radius of curvature, one findsmq

= rB0

v= rB0B

E.

Measuring r, B0, B,E, one can determine mq

.

Physics 201 – p. 25/30

Magnetic Fields

Some applications: Mass spectrometer

After passing through a velocity selector, theparticles go through a second region withuniform magnetic field B0 in the samedirection as the one in the velocity selector.They bend around and using the expressionfor the radius of curvature, one findsmq

= rB0

v= rB0B

E.

Measuring r, B0, B,E, one can determine mq

.

Physics 201 – p. 25/30

Magnetic Fields

Some applications: Mass spectrometer

Physics 201 – p. 26/30

Magnetic Fields

Magnetic force on wire with current

A wire with current contains flowing electrons.

In the presence of a magnetic field, eachelectron experiences a magnetic force~F = −e~velectron × ~B

Notice that by convention, the direction of thecurrent is opposite to that of the electrons i.e.~v+ = −~velectron. Equivalent to ~F = e~v+ × ~B i.e.force on “positive” charges moving in theopposite direction to that of the electrons.

Physics 201 – p. 27/30

Magnetic Fields

Magnetic force on wire with current

A wire with current contains flowing electrons.

In the presence of a magnetic field, eachelectron experiences a magnetic force~F = −e~velectron × ~B

Notice that by convention, the direction of thecurrent is opposite to that of the electrons i.e.~v+ = −~velectron. Equivalent to ~F = e~v+ × ~B i.e.force on “positive” charges moving in theopposite direction to that of the electrons.

Physics 201 – p. 27/30

Magnetic Fields

Magnetic force on wire with current

A wire with current contains flowing electrons.

In the presence of a magnetic field, eachelectron experiences a magnetic force~F = −e~velectron × ~B

Notice that by convention, the direction of thecurrent is opposite to that of the electrons i.e.~v+ = −~velectron. Equivalent to ~F = e~v+ × ~B i.e.force on “positive” charges moving in theopposite direction to that of the electrons.

Physics 201 – p. 27/30

Magnetic Fields

Magnetic force on wire with current

In a segment of a wire with cross section Aand length L, the number of charged particleswith a drift speed vd is nAL where n is thenumber of particles per unit volume.

The magnetic force on the wire is~F = (q~vd × ~B)nAL

With the current being I = nqvdA, one gets~F = I~L × ~B~L: vector in the direction of the current andwith magnitude L.

Physics 201 – p. 28/30

Magnetic Fields

Magnetic force on wire with current

In a segment of a wire with cross section Aand length L, the number of charged particleswith a drift speed vd is nAL where n is thenumber of particles per unit volume.

The magnetic force on the wire is~F = (q~vd × ~B)nAL

With the current being I = nqvdA, one gets~F = I~L × ~B~L: vector in the direction of the current andwith magnitude L.

Physics 201 – p. 28/30

Magnetic Fields

Magnetic force on wire with current

In a segment of a wire with cross section Aand length L, the number of charged particleswith a drift speed vd is nAL where n is thenumber of particles per unit volume.

The magnetic force on the wire is~F = (q~vd × ~B)nAL

With the current being I = nqvdA, one gets~F = I~L × ~B~L: vector in the direction of the current andwith magnitude L.

Physics 201 – p. 28/30

Magnetic Fields

Magnetic force on wire with current

Its magnitude isF = ILB sin θθ: Angle between the magnetic field and thecurrent.

It can be shown that the net force on a closedcurrent loop in a uniform magnetic field iszero.

Physics 201 – p. 29/30

Magnetic Fields

Magnetic force on wire with current

Its magnitude isF = ILB sin θθ: Angle between the magnetic field and thecurrent.

It can be shown that the net force on a closedcurrent loop in a uniform magnetic field iszero.

Physics 201 – p. 29/30

Magnetic Fields

Magnetic force on wire with current

Physics 201 – p. 30/30