General Case Magnetic Field - McMaster Physics and Astronomy

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Magnetic Forces and Torques
Review:
Charged Particle in an external field:



F  qv  B
Straight wire of length L with current I in a
uniform external field B:
F  I LB
Note: L points in direction of positive current flow
Force on a current-carrying wire (general case)
If B not uniform, and/or wire not straight: the
force dF on a short segment of vector length dL is
dF = IdL x B
The total force on the wire is:
I
dF
B
Segment of
length dL
dL
I
F   I dL  B
along
wire
Example 1
Find the force on:
y
x
x
x
x
I R
x
x
a) The straight wire
b) The semicircular wire
c) The whole circuit
x

x
x
x
B (uniform)
For (b): start with force dF due to an
infinitesimal piece, and do the integral.
Solution
Theorem: For any closed current loop in a uniform
magnetic field,
Total magnetic force on the loop = 0
Proof:
F   I dL  B
 I  dL  B
 
(if B is a constant vector)
But:  dL  0 for a closed loop, so F = 0
Torque:
Although there is no net force on a circuit in a
uniform field, there may be a net torque. The torques
due to equal and opposite forces applied at different
locations do not necessarily cancel.
We can calculate the torque directly for a rectangular
loop. There is also a simple rule, which actually applies
to a plane loop of any shape. First we need to define
the “magnetic dipole moment” (a vector) for a current
loop.
Magnetic Dipole Moment


(or “magnetic moment”)
Define the magnetic moment of a small current loop by


  IA

(A  " vector area" of loop)
Area
of loop

current
Note the right-hand
rule! – fingers follow I,
thumb points in μ
( surface)
Torque on a Current Loop (Uniform B)
Example: a rectangular loop
B
A


h
B
C
D
w
Forces:
FAD= I h B
FAB= I w B sinθ
x
FBC= I h B
FCD= I w B sinθ
Top view:


FBC= I h B
I
x

B
B
I into the page
A
FAD = I h B
w•sinθ
Torque (about any pivot; e.g., at A)
 IhB  w sin 
 ( Ihw) B  sin 
 ( I  area )  B  sin     B
B
The torque on a current loop with
magnetic dipole moment  in a
uniform external field B is:



   B
Where μ = IA
Example 2
Find the torque on a flat, horizontal, circular coil due to the
magnetic field of the Earth.
B
north
B
50o
0.5 x 10-4 T
Circular loop, 2 turns, R = 1 m, I = 20 A,
CCW (from above)
Find: Torque (magnitude and direction)
Example 2
B
B
50o
I
Which direction does the torque vector point in for a flat
horizontal coin on the surface of earth ?
Example 3
Find the torque on a flat, horizontal, circular coil due to the
magnetic field of the Earth.
B
north
B
50o
0.5 x 10-4 T
Circular loop, 2 turns, R = 1 m, I = 20 A (ccw from above)
Example 4
Consider an electron orbiting a proton and maintained in a fixed
circular path of radius R=5.29x10-11 m by the Coulomb force.
Treating the orbiting charge as a current loop, calculate the
resulting torque when the system is in an external magnetic field
of B=0.4 T directed perpendicular to the magnetic moment of the
orbiting electron.
Summary
Charged Particle:

 
F  qv  B
Wire:
dF  I dL  B
or
Straight wire, Uniform B:

 
F  IL  B
Torque:
  
   B
F   I dL  B

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