Download Lesson 18 (1) Force on a Current Loop in a Uniform Magnetic field

Lesson 18
(1) Force on a Current Loop in a Uniform Magnetic field
For an arbitrary current loop in a uniform magnetic field, the total force is
F=
ò Id
´B= I
( ò d )´B = 0
The loop will not undergo translational acceleration. Instead, it can turn.
(2) Torque
Consider a force F acting on a point P of an object. Choose any point O in space and
define the vector r to be the displacement vector from O to P. We define
Torque of F about O t = r ´ F
If a number of forces act on an object, and the sum
of the forces is zero, the total torque is
independent of the choice of the point O.
Proof: Let another point O be chosen to calculate
torque, and denote the displacement vector from


O to O be a , that from O to the point P by r  .
Then

 
  
  
 




r

F

r

a

F

r

F

a

F

r



   F
The torque vector for a force couple (two forces of equal
magnitude and opposite directions) is shown. The
magnitude of the torque is

  F d
where d is the shortest distance between the lines of
actions of the forces.
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The relation between the sense of rotational motion caused by
the torque and the direction of the torque vector is given by a
right hand rule: holding a fist with your right hand with thumb
stuck out pointing in the direction of the torque vector, the
fingers indicate the sense of rotational motion. (actually
acceleration)
(3) Torque on a Current Loop and Magnetic Moment
Consider a rectangular loop ABCD with width w and length  as shown:
w
B
A
N
S

C
D
The top view of the loop and magnet is shown below:
FDA
A
w
θ
B
FBC
The forces on the four sides of the loop are
FAB
FCD
FBC
FDA
 IwB sin  2   
 IwB sin  2   
 IB
down
 IB
up
out of paper
int o paper
and they cancel in pairs.
While the first two forces have no effect on the motion of the loop, the last two
forces form a force couple that tends to rotate the loop in the clockwise direction.
The magnitude of the torque of this force couple is
2
  FBC w sin   IwB sin   IAB sin 

where A  w is the area of the loop. Using the right hand rule, the direction of  is

into the paper. We now define the magnetic moment  of an N-turn current loop of
area A with current I as follows:

  NIA
Direction given by right  hand  rule
Here the right hand rule is as follows: On right hand
fist with stuck-out thumb, the fingers represent the
direction of current in the loop, and the thumb gives

the direction of  .
In terms of the magnetic moment, the torque on the rectangular wire loop can be
found from



  B
The action of a magnetic field on the magnetic moment of a current loop is similar to
the action of electric field on an electric dipole moment vector, which points from
the negative to the positive charge. At equilibrium, the torque on a current loop is


zero. This occurs when  is parallel or anti-parallel to B . Stable equilibrium
corresponds to the former case.
(4) Proof of the torque on current loop formula in general
Proof: Consider an arbitrary current loop on the x-y plane in the presence of a


uniform magnetic field B  Bˆj . The force on an element d  is



 
dF  Id   B  I dxiˆ  dyˆj  Bˆj  IBdx kˆ
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The torque about the origin of this force is



 
d  r  dF  xiˆ  yˆj  IBdx kˆ   IBxdxˆj  IBydx iˆ
The total torque is






   d  IB   xdxˆj   ydxiˆ  IBA iˆ   IABkˆ  ˆj    B
Here we have used the following integrals:
 xdx  0
 ydx  A
(4) The electric Motor
The operation of a DC motor relies on the rotation of a wire loop inside a magnetic
field that arises when a current is sent through the wire.
Visit the following
website for a simulation of a dc motor.
http://www.ul.ie/~gaughran/flynn/dcmotor.html
The directions of various vectors and the sense of rotation of the loop are sketched
as shown. To keep rotating in the same direction, the direction of the current must
be reversed in a half cycle. For example, the current on the left arm of the loop in the
diagram must always point out of the paper even though after half a cycle, the arm is
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replaced by the other one. This reversal of current is accomplished by the spit-ring
commutator.
(5) Energy of a Current Loop in a Magnetic Field
This energy is calculated from the work required to rotate the current loop. For that
we first need the expression for the work done by a torque on an object.
For simplicity, let the object be a rod of length L , and a force F acts perpendicularly
to the rod at one end, so that the torque about the other end is
t = FL
turning the rod in the counter-clockwise direction. Let q be
the angle between the rod and the x-axis, increasing in the
counter-clockwise direction. When the rod rotates through
the angle dq , the distance travelled by the end of the rod
where the force acts is Ldq , so that the work done by this
force is
dW = FLdq = t dq
and is also the work done by the torque.
If the rod is replaced by the magnetic moment vector of a current loop as shown,
and the magnetic field is in the positive x-direction, the torque has magnitude
m Bsinq and its direction is into the paper. It therefore tends to rotate the magnetic
moment vector clockwise, and we can write
t = -m Bsinq
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The increment of energy of the loop is equal to the work done against this torque:
dU = -t dq = m Bsinq dq
Upon integrating,
U = -m Bcosq + C
Choosing as a reference q = p 2 to be where U = 0 , the constant C = 0 . The energy is
U = -m Bcosq = -m × B
The dependence of U on q is sketched below. The energy is minimum at q = 0 , when
the magnetic moment is parallel to the magnetic field.
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