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PHYSICS II
MAGNETOSTATICS
WHY SHOULD YOU CARE?
Practical uses of magnetism:
Magnetic resonance imaging (MRI)
Magnetic hypothermia
Magneto-optic screening
A neurosurgeon can:
Guide a pellet through brain tissue to inoperable tumors.
Pull a catheter into place.
Implant electrodes while doing little harm to brain tissue.
MAGNETOSTATICS
Electrostatics:
All sources charges are static, meaning they aren’t moving..
Produces an electric field 𝐸.
Magnetostatics:
All charges are moving, but the movement is
static, meaning it’s a constant rate of movement.
Produces a magnetic field 𝐵.
OBSERVATIONAL EXPERIMENT
Imagine the following:
Two wires hand from the ceiling, a few
centimeters apart.
I turn on a current.
Do the wires…
Repel each other?
Attract each other?
Remain where they are?
OBSERVATIONAL EXPERIMENT
a) Currents in
opposite directions
repel.
b) Currents in same
directions attract.
TAKE-AWAY
This is our first encounter with the magnetic force.
Stationary charge produces only the electric field 𝐸.
Moving charge, in addition, produces the magnetic field 𝐵.
MAGNETIC FIELDS
As with the electric field, we can represent the magnetic
field by means of drawings with magnetic
field lines.
These field lines always go from
north to south.
MAGNETIC FIELDS
Our earth has
magnetic north
and south.
The field protects
use from cosmic
particles.
MAGNETIC FIELDS
If you hold up a compass outside, the
needle will point to the north.
But if you hold a compass up to a
current-carrying wire, the result is
peculiar.
The compass will not point toward
the wire, nor away from it, but
rather it circles around the wire.
MAGNETIC FIELDS
How can such a field create an
attraction between two wires?
On the second wire…
The magnetic field is into the page.
The velocity of the charges is upward.
The resulting force is to the left.
A very strange law is needed to
account for these directions.
MAGNETIC FORCE
If you’re perceptive, you’ll notice the odd behavior follows what is
knows as the Right-Hand Rule.
MAGNETIC FORCE
MAGNETIC FORCE
Mathematically, this is described by a cross product.
𝐹𝐵 = 𝑞 𝑣 × 𝐵
A cross product simply says that 𝐹𝐵 is perpendicular to both 𝑣 and 𝐵.
The magnitude of the magnetic force is simply
𝐹𝐵 = 𝑞𝑣𝐵sin𝜃
Where 𝐹𝐵 is in newtons, 𝑞 is in coulombs, 𝑣 is in m/s, and 𝐵 is in the
unit called tesla (T).
NOTATION
The vector 𝐵 is sometimes seen in the perspective shown in the figures.
THINK!
EXAMPLE 1
An electron is accelerated through ∆𝑉 = 2400 V from rest and then
enters a uniform 𝐵 = 1.70 T magnetic field. What is
a.) the maximum value of the magnetic force this particle can
experience.
b.) the minimum value of the magnetic force this particle can
experience.
This will be done on the board.
CHARGED PARTICLE IN UNIFORM 𝐵-FIELD
A positively charged particle is
injected into a magnetic field 𝐵
with a velocity 𝑣 exactly
perpendicular to the field.
The particle is put into circular
motion.
What’s its radius of motion?
Let’s apply Newton’s second law.
CHARGED PARTICLE IN UNIFORM 𝐵-FIELD
𝐹 = 𝑚𝑎
⇒
⇒
𝐹𝐵 = 𝑚𝑎
𝑣2
𝑞𝑣𝐵sin𝜃 = 𝑚𝑎 = 𝑚
𝑟
Centripetal acceleration
Since 𝑣 ⊥ 𝐵, the angle 𝜃 = 90°, which gives sin 90° = 1.
⇒
⇒
𝑣2
𝑞𝑣𝐵 = 𝑚
𝑟
𝑚𝑣
𝑟=
𝑞𝐵
EXAMPLE 2
A proton (𝑞 = 1.60 × 10−19 C) is moving in a circular orbit of
radius 14 cm in a uniform 0.35 Tesla magnetic field
perpendicular to the velocity of the proton. Find the speed of
the proton.
This will be done on the board.
EXAMPLE 3
An electron in a television picture tube
moves toward the front of the tube with
a speed of 8.0 × 106 m/s along the 𝑥axis. Surrounding the neck of the tube
are coils of wire that create a magnetic
field of magnitude 0.025 T, directed at
an angle of θ = 60° to the 𝑥-axis and
lying in the 𝑥𝑦 plane. Calculate the
magnetic force on the electron.
This will be done on the board.
LORENTZ FORCE
If a charged particle is in the presence of an electric field 𝐸 and a
magnetic field 𝐵, then the total force (Lorentz force) is
𝐹 = 𝑞𝐸 + 𝑞𝑣 × 𝐵
Let’s take a look at an unintuitive trajectory created by the
combination of these forces.
CYCLOID MOTION
Suppose that 𝐵 point in the 𝑥-direction and 𝐸 in the 𝑧-direction. A
positively charged particle at rest is released from the origin. What
path will it follow?
EXAMPLE 4: VELOCITY SELECTOR
If the speed of a proton is properly
chosen, the proton will not be
deflected by these crossed electric
and magnetic fields. What speed
should be selected in this case?
This will be done on the board.
MASS SPECTROMETER
A mass spectrometer separates ions according to
their mass-to-charge ratio.
The equation for the radius of a charge particle in
a 𝐵-field is
𝑚𝑣
𝑟=
𝑞𝐵2
And the velocity of an undeflected particle in a velocity
selector is
𝐸
𝑣=
𝐵1
Therefore the mass-to-charge ratio here is
𝑚 𝑟𝐵2 𝐵1
=
𝑞
𝐸
𝐵-FIELD AND CURRENT
𝐵-FIELD AND CURRENT
The equation for the magnetic force created by a magnetic field 𝐵
on a current-carrying wire with current 𝐼 and length 𝐿 is
𝐹𝐵 = 𝐼𝐿 × 𝐵
The magnitude of the magnetic force is therefore
𝐹𝐵 = 𝐼𝐿𝐵sin𝜃
TORQUE ON CURRENT LOOP
TORQUE ON CURRENT LOOP
𝑏
𝑏
𝜏 = 𝐹2 sin𝜃 + 𝐹4 sin𝜃
2
2
𝑏
𝑏
= 𝐼𝑎𝐵 sin𝜃 + 𝐼𝑎𝐵 sin𝜃
2
2
= 𝐼𝑎𝑏𝐵sin𝜃
= 𝐼𝐴𝐵sin𝜃
EXAMPLE 5
A rod of mass 0.720 kg and radius
6.00 cm rests on two parallel rails that
are 𝑑 = 12.0 cm apart and 𝐿 =
45.0 cm long. The rod carries a current
of 𝐼 = 48.0 A in the direction shown and
rolls along the rails without slipping. A
uniform magnetic field of magnitude
0.240 T is directed perpendicular to the
rod and the rails. If it starts from rest,
what is the speed of the rod as it leaves
the rails?
TORQUE ON CURRENT LOOP
The general equation for the torque on a current loop placed in a
uniform magnetic field 𝐵 is
𝜏 = 𝐼𝐴 × 𝐵
where the area vector 𝐴 is perpendicular to the
plane of the loop.
Use the Right-Hand Rule to determine the
direction of 𝐴.
The magnitude of this equation is
𝜏 = 𝐼𝐴𝐵sin𝜃
MAGNETIC DIPOLE MOMENT
The magnetic dipole moment 𝑚 is defined, in general, by the
following relation:
𝜇=𝐼
𝑑𝐴
For a flat current loop, 𝐴 is the area enclosed by the loop, so
𝜇 = 𝐼𝐴
Magnetic dipole
moment of a current
loop.
MAGNETIC DIPOLE MOMENT
If a coil of wire contains 𝑁 loops of equal area, the magnetic dipole
moment becomes
𝜇 = 𝑁𝐼𝐴
The torque on a current-carrying wire loop in magnetic field 𝐵
becomes
𝜏 =𝜇×𝐵
THINK!
Rank the magnitudes of the torques acting on the rectangular loops (a), (b), and
(c) shown edge-on in the figure from highest to lowest. All loops are identical and
carry the same current.
Rank the magnitudes of the net forces acting on the rectangular loops from
highest to lowest.
EXAMPLE 6
A rectangular coil of dimensions 5.40 cm × 8.50 cm consists of 25
turns of wire and carries a current of 15.0 mA. A 0.350 T magnetic
field is applied parallel to the plane of the coil. What is the torque on
the coil?
This will be done on the board.
EXAMPLE 7
Find the magnetic dipole
moment 𝜇 of a “bookendshaped” loop shown in the
figure. The height and width
are 𝑑, and the length is 𝑙. The
entire loop carries a current
𝐼.
This will be done on the
board.