Download Magnetic Force on a Current-Carrying Conductor

Magnetic Force on a Current-Carrying
Conductor
AP Physics C
Montwood High School
R. Casao
• Just as a force is exerted on a single
charged particle when it moves through a
magnetic field, a current-carrying wire also
experiences a force when placed in a
magnetic field.
• A current consists of many charged
particles in motion; the resultant force on
the wire is the sum of the individual forces
on the charged particles constituting the
current.
• The magnetic field B is
directed out of the
page.
• Figure a: no current
flows in the wire, so
there is no force acting
on the wire.
• Figure b: for the
current I going from
bottom to top, the
magnetic force is
directed to the right.
• Right-hand rule:
fingers of right hand in
direction of the velocity
of the current (bottom
to top); palm faces out
away from page in
direction of magnetic
field B; thumb points to
the right.
• Figure c: for the
current I going from
top to bottom, the
magnetic force is
directed to the left.
• Right-hand rule:
fingers of right hand in
direction of the velocity
of the current (top to
bottom); palm faces
out away from page in
direction of magnetic
field B; thumb points to
the left.
• Consider a straight
segment of wire of
length l and crosssectional area A
carrying a current I in
a uniform magnetic
field B.
• The magnetic force
on a charge q moving
with a drift velocity vd
is given by: q·v x B.
• The force on the charge
carriers is transmitted to
the wire through
collisions with the atoms
making up the wire.
• The total force on the
wire is found by
multiplying the force on
one charge by the
number of charges in the
segment.
• The volume of the
segment is A·l.
• The number of charges in the segment is n·A·l,
where n is the number of charges per unit
volume.
• Total magnetic force on the wire of length l:
Fmag = (q·vd x B)· n·A·l
• Remember that current I = n·q·vd·A. Therefore:
Fmag = I·(l x B), where l is a vector in the
direction of the current.
– The magnitude of l equals the length of the wire
segment.
• This equation only works for a straight segment of wire
in a uniform external magnetic field.
• The magnetic field produced by the current itself is also
ignored. The wire cannot produce a force on itself.
• Consider a wire of
uniform cross-sectional
area in an external
magnetic field, as
shown in the figure.
• Dividing the wire into
small segments ds that
can are approximately
straight, we can
describe the magnetic
force on a very small
segment ds in the
presence of a
magnetic field B as:
dFmag
 
 I  d s x B
• dFmag is directed out of the page.
• The magnetic field B can be defined as a
measurable force on a current element.
– The force is maximum when B is
perpendicular to the current element
– The force is zero when B is parallel to the
current element.
• To get the total force F, integrate from one
end of the wire (a) to the other end of the
wire (b).
 
b
 dF  I   d s x B
mag
a
• Conclusion: the magnetic force on a
curved current-carrying wire in a uniform
magnetic field is equal to that on a straight
wire connecting the end points and
carrying the same current.
• Case I: for a curved wire carrying a
current I located in a uniform external
magnetic field B.
• The constant magnetic field B can be
pulled out in front of the integral:
 dF
mag
 b 
 I  B  a d s

• The quantity a ds represents the vector
b
sum of all the displacement elements from
a to b (the length of the conductor, s).
• Conclusion:
 
 
F  I  s x B  I  s  B  sin θ
• Case II: consider a closed loop carrying a
current I placed in a uniform external magnetic
field B, pull B out in front of the integral.
 dF
mag
 b 
 I  B  a d s
• The sum of the
displacement vectors
ds is zero for the
closed loop because
the vectors form a
closed loop.

 d s  0 , the force Fmag = 0 N.
• Since
• Conclusion: the total magnetic force on any
closed current loop in a uniform magnetic field is
zero.
• Problem Example 29-2: Force on a Semicircular
Conductor
• A wire bent into the shape of a semicircle of
radius R forms a closed circuit and carries a
current I. The circuit lies in the xy plane, and a
uniform magnetic field is present along the
positive y axis. Find the magnetic forces on the
straight portion of the wire and on the curved
portion.
• Straight portion of wire:
– Divide the length of the straight portion into equal
elements of length ds.
– The angle between each equal element of length ds
and the magnetic field B is 90º.
– Using the right hand rule and crossing ds into B
indicates that the direction of the resulting magnetic
force is out of the page.
• The net force on the straight portion of the wire
is the sum of the force contribution from each
element of length ds. Integrating from the left
end L to the right end R:
 R 
L dFmag  I  B  L d s
 
Fmag  I  B  s

s  2R

Fmag  I  B  2  R

Fmag  2  I  B  R
R
L
R
Direction: out of the page.
• Semicircle portion:
– Divide the semicircle into
small elements of length ds
from the right end R to the
left end L.
– Each element of length ds
is a different angle q from
the vertical magnetic field
B.
– Divide the element of
length ds into a component
parallel to the magnetic
field B and perpendicular
to the magnetic field B.
– Remember that the magnetic
force is zero for vector
components parallel to the
magnetic field and maximum for
vector components
perpendicular to the magnetic
field.
– The equation for the
perpendicular components is
ds·sin q.
– The angle q will vary from 0 rad
to p rad for each element of
length ds along the semicircular
arc.
– arc length equation: s = R·q.
• For each element of
length ds:
 
dFmag  I  B  d s  sin θ


s  R θ
d s  R  dθ

dFmag  I  B  R  sin θ  dθ
• To get the total force
on the semicircular
portion, integrate from
0 rad to p rad:

0 dFmag  0 I  B  R  sin θ  dθ

p
Fmag  I  B  R  0 sin θ  dθ

π
Fmag  I  B  R   cos θ  0

Fmag  I  B  R   cos π - cos 0 

Fmag  I  B  R    1 - 1


Fmag  I  B  R  1  1  2  I  B  R
π
π
• The direction for the force on the
semicircular portion of the wire is into the
page.
• Combine the two results. Take out of the
page as the positive direction (the positive
z-axis) and into the page as the negative
direction (the negative z-axis).
Fmag  2  I  R  B  2  I  R  B  0 N
This result shows that the net force on a
closed loop is 0 N.
• The direction of the magnetic field
due to a current carrying element is
perpendicular to both the current
element ds and the radius vector
rhat.
• The right hand rule can be used to
determine the direction of the
magnetic field around the current
carrying conductor:
– Thumb of the right hand in the
direction of the current.
– Fingers of the right hand curl
around the wire in the direction of
the magnetic field at that point.
• The magnetic field lines are
concentric circles that surround
the wire in a plane perpendicular
to the wire.
• The magnitude of B is constant
on any circle of radius a.
• The magnitude of the magnetic
field B is proportional to the
current and decreases as the
distance from the wire increases.