Download Magnetic Field Inside a Solenoid

Magnetic Field of a Current Carrying Wire
What we are about to derive can aslo be applied to any current type scenario, say a beam of
charged particles (protons, alpha particles and electrons immediately come to mind). Take note,
however, that when we say current in this context we imply conventional current, a.k.a. a flow of
positive charge. Remember that the direction of the B-field resulting from electron flow would
be opposite what your RHR gives you.
For a wire with current flow into the page:
We place an “amperean loop” around the cross-section of the wire. The shape of the loop is
arbitrary, but intended to form a 2-D path on which the intensity of the field is uniform, or at
least easy to determine (see the next problem as an alternative). For a long straight wire, the Bfield makes uniform, revolving, concentric cylinders around the wire. To find the field intensity
B at point P a distance d away, a circular shape makes sense because every point on the circle
has the same B-field intensity, being the same distance from the wire. Note that we could easily
substitute x, r, or L or any special variable for the distance from the center. I chose d for this
problem as it relates to the two wire problem further down this page.
Applying Ampere’s Law, we have to solve the path integral (2D) adding up all the bits of B as
we move around the circle from start to finish: that is from zero to 2*pi (the circumference of the
circle!):
 B  dL  oiencl Ampere’s Law
B is constant at all points on the
B  dL  oiencl
path (the circle)
The integral of the path is the
B(2 d )  oi
circumference!
o i
B
B-field due to a long, straight
2 d
wire at a distance D from the wire.
Inside the wire:
Magnetic Field Inside a Solenoid
Inside the solenoid the B-field is directed to the right according to the right-hand-rule when the
current i is out of the page on top and into the page at bottom, as shown above.
Use Ampere’s Law.
1) Establish an Amperean loop that will either give you an identical B-field at every point on the
loop (this symmetry allowing B to factor out of the integral) OR create a loop that will give you
identical B-field on each part, allowing you to break it into several integrals with B of constant
intensity for each part.
In this case any rectangle like ABCDA will work well (assuming a square is fine). The
B-field at each side is unique and of a constant value.
 B  dL  oiencl

B
A

B  dL 
C
B
B  dL 

D
C
A
B  dL   B  dL  oiencl
D
Recall that the dot product is the scalar projection of one vector onto another. This is like the
shadow of the second vector onto the first.
The magnitude of the dot product is given by the formula
A B  A B cos
where theta is the angle between vector A and vector B. This product is maximized when vector
A and vector B are parallel (cos 0o = 0) and zero when A and B are perpendicular (cos 90o = 0).
Therefore, for the form of Ampere’s Law above, several terms cancel:

B
A
B  dL  0 

D
C
B  dL  0  oiencl
Now consider that since the region CD is outside the solenoid, we know the field to be
negligible there. So

B
A
B  dL  0  0  0  oiencl
Now we need only define the enclosed current and we can find the B-field inside the solenoid.
Most of the time the solenoid is defined as carrying current i, with n-turns of wire per unit length.
The n variable is important as are its units. Consider the enclosed current of any number of sideby side wires inside the Amperean loop.
iencl = (# of wires)(current i in each wire)
Now look at the variable n
n
iencl
# wires
length
 (# wires )(current in each)
iencl  nhi
Check the units:
# wires
 length  current
length
and we can fairly easily define iencl.
Back to Ampere’s Law:

B
A
B  dL  oiencl
B
B  dL  oiencl
A
Path length from A to B is h, substitute i encl
Bh  o nhi
B  o ni
Note that the intensity of the field does not depend on the dimensions of the solenoid. This
means that B is uniform everywhere inside!
PARTICLE IN A MAGNETIC “BOTTLE” (2D)
A proton (q=+1.9e-19, m=1.67e-27) enters a region of uniform magnetic field with intensity
B=3.0e-3 Tesla and a velocity of 1000 m/s in the y direction.
a)
b)
c)
d)
e)
What path will the proton follow in the field?
Find the magnetic force on the particle.
Find the acceleration of the particle due to the field.
Determine the radius of the circular path.
Qualitatively describe the behavior of the particle if the original velocity vector was
instead v = 50i + 1000j m/s.
Magnetic Force from 2 long wires w/ opposing, currents.
Strategy: Find the force of wire A’s field on wire B (a distance d away) and then find the
force of wire B’s field on wire A (a distance d away). Add the vectors.
Field of wire A at distance d:
 B  dL   i
B  dL   i
o encl
o encl
B(2 d )  oi
B
o i
2 d
field direction is to the bottom of the page by RHR.
The magnetic force on a wire carrying a current in a B field is given by F=iLxB, which
comes from F=qvxB. In this equation the I refers to the current in the wire that experiences
the force. In this problem the magnetic field is produced by a second I…gets confusing.
F  iL  B
i
F  (2i )( L)( o ) sin 90o
2 d
2
i Lo
F
d
This is the force of wire A’s field on wire B. Direction of the force is to the right by RHR.
If the currents were the same (they are NOT) we would simply double this magnitude,
realizing opposite direction the wires would repel each other. While the latter is still true, we
must still find the magnitude of the B-field from wire B at a distance d (as we did for A,
above) and substitute into the force equation.
Then add both forces (since they are both “repelling” forces). Indicate direction: the wires
get pushed apart.
Variations of the problem include wires carrying currents in the same direction or a simple
geometric arrangement of wires (say three in a triangle) where the B-field from each wire
exherts a force on both other wires…need to find resultant vectors of the two forces on each
wire.
Torque on a Current Loop:
This is the basic principle behind an electric motor. A loop, or many loops of current
carrying wire are called an armature. When the armature is placed into an external B-field
(usually created by solenoid electromagnets) the armature will begin to rotate. That is, the
magnetic force on the loop will be in opposite directions on each side, producing a torque.
Find the external field (made by the magnets).
Find the direction of current in each straight section of wire.
Use the RHR to determine the direction of magnetic force on each straight section of wire
(some will be zero).
Recall that torque is the cross product of the force and radius vector. In the past we have
often assumed that F and r are at right angles of each other, maximizing the cross-product.
More accurately we should write:
  rF
  F r sin 
F and r will always be at a right angle so the sine function drops out
 F r
Recall, F=iLxB, Substitute for the magnetic force
  iLBsin r
  iLBrsin
where  is the angle between L (the wire) and B (the field)
If you have many turns (loops) of wire in the armature, torque is produced on each loop, thus
simply multiply by the number of loops, n, to get the total torque.
  niLBrsin