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Slide 1
AP Physics C Magnetic Fields and Forces
Slide 2
Currents Set up Magnetic Fields First Right-Hand Rule Hans
Christian Oersted (1777-1851)
Slide 3
Right-Hand Rule for Magnetic Fields x: into the page or away
from you out of the page or towards you
Slide 4
Magnetic Field of a Current Loop:
Slide 5
Another look a Magnetic Fields of a Current Loop:
Slide 6
Electron as a Magnetic Dipole The electron spins on its axis,
giving rise to a electron current in the direction of rotation. The
electron is like a magnetic dipole, a miniature magnet, with a
north end and a south end.
Slide 7
Magnetic Spin Dipoles in Iron
Slide 8
Dual polarity Cutting a magnet in half will not isolate a
single north or south. One magnet becomes two, then four, and so
on. This process will never end; even when the last electron spin
dipole is reached, it cannot be cut to reveal a single north or
single south pole.
Slide 9
Magnetic Fields of a Bar Magnet & the Earth B-field of bar
magnet is similar to the Earth's magnetic field. B-field lines
leave north face, enter at south face. Convection currents inside
the earth set up magnetic field.
Slide 10
Bar Magnets If magnetic dipole loops are oriented the same on
neighboring faces, the magnets attract. North is attracted to
south, and is repelled by north.
Slide 11
Compass Needle is a Magnet: It Aligns with the B-Field
Slide 12
Magnetic Force on Current-Carrying Wire F = ILB sin The
direction of the force on the wire may be determined by a second
right-hand rule, a right hand rule for magnetic force. The other
right-hand rule gave the direction of the magnetic field B.
Slide 13
Using the Right-Hand Rule To determine the direction of the
magnetic force acting on the current-bearing wire
Slide 14
Using the RHR: In which direction, if any, will the metal rod
be deflected?
Slide 15
Sample Problem A wire 2.80 m in length carries a current of
5.00 A in a region where a uniform magnetic field has a magnitude
of 0.390 T. Calculate the magnitude of the magnetic force on the
wire if the angle between the magnetic field and the current is
60.0 degrees.
Slide 16
Sample Problem A thin, horizontal copper rod 1.0 m long and has
a mass of 50 g. What is the minimum current in the rod that will
cause it to float in a horizontal magnetic field that is
perpendicular to the rod of 2.0 T?
Slide 17
Torque on a Current Loop What is the net force on the current
loop? What is the net torque on the loop?
Slide 18
Torque on the Current Loop At what angle is the torque a
maximum value? At what angle is the torque a minimum value?
Slide 19
Sample Problem A circular loop of radius 2.0 cm contains 50
turns of tightly wound wire. If the current in the windings is 0.30
A and a constant magnetic field of 0.20 T makes an angle of 25
degrees with a vector perpendicular with the loop, what torque acts
on the loop?
Slide 20
DC Motor
Slide 21
Moving Charges in a B-Field: Electric force can be parallel to
direction of velocity, but the magnetic force is always
perpendicular to the velocity vector.
Slide 22
Magnetic Force on Moving Charges RHR F = qvB sin
Slide 23
Magnetic Force on Moving Charges If the velocity v is parallel
to the magnetic field B, the magnetic force is zero because sin =
0.
Slide 24
Magnetic Force on Moving Charges What is the direction of the
magnetic force on the moving charge in each situation?
Slide 25
Magnetic Force on Moving Charges What is the direction of the
force F, if any, in each case?
Slide 26
Charges move in a circular path:
Slide 27
Circular Paths in Magnetic Fields
Slide 28
Circular Motion in B-Field Right Hand Rule for Force Fingers
point in direction of magnetic field B. Thumb points in direction
of the velocity vector v. Palm shows the direction of the force
F.
Slide 29
Mass Spectrometer F = qvB sin = 90 deg F = qvB F = ma a = v 2
/r F = mv 2 /r qvB = mv 2 /r m = qBr/v
Slide 30
Sample Problem A singly charged positive ion moving at 4.6 x 10
5 m/s leaves a circular track of radius 7.94 mm along a direction
perpendicular to the 1.80 T magnetic field of a bubble chamber.
Compute the mass (in amus) of this ion, and identify it from that
value.
Slide 31
Velocity Selector & Mass Spectrometer In the velocity
selector, the E-force and the B-force are equal and opposite, so
that, qE = qvB. Therefore, v = E/B. In 1897, J. J. Thomson used
this set-up to determine the mass to charge ratio for
electrons.
Slide 32
Sample Problem The electric field between the plates of a
velocity selector is 2500 V/m, and the magnetic field in both the
velocity selector and the deflection chamber has a magnitude of
0.0350 T. Calculate the radius of the path of a singly charged ion
have a mass of 2.18 x 10 -28 kg.
Slide 33
Electron Beam in a B-Field Electrons are deflected downward.
What is the direction of the magnetic field B?
Slide 34
Magnetic Flux Magnetic flux is the product of the average
magnetic field times the perpendicular area that it
penetrates.magnetic field
Slide 35
Magnetic Flux Illustrations The contribution to magnetic flux
for a given area is equal to the area times the component of
magnetic field perpendicular to the area. For a closed surface, the
sum of magnetic flux is always equal to zero (Gauss' law for
magnetism).magnetic fluxGauss' law for magnetism
Slide 36
Gausss Law for Magnetism The net magnetic flux out of any
closed surface is zero. This amounts to a statement about the
sources of magnetic field. For a magnetic dipole, any closed
surface the magnetic flux directed inward toward the south pole
will equal the flux outward from the north pole. The net flux will
always be zero for dipole sources.magnetic flux
Slide 37
Sample Problem A cube of edge length 2.50 cm is positioned so
that it is position with one corner at the origin, one face in the
xy- plane, one face in the yz-plane and one in the xz-plane. A
uniform magnetic field given by B = (5.00i +4.00j +3.00k) T exists
throughout the region. Calculate the flux through the face that is
parallel to the yz- plane. What is the total flux through the six
faces?