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
vd
A
5. Magnetic forces on current

 
F  qv  B
l
n
N
lA
N  nlA
 
lvd  l vd
nqAvd  I


 - angle between vd and B
 
 
F  Ftot  Nqvd  B
 
 nlAqvd  B
 
 nqAvd l  B
  
F  Il  B
 

F  Il  B
F  IlB sin 
F  IlB sin 
Example: A straight wire carrying a current is placed in a region containing a
magnetic field. The current flows in the +x direction. There are no magnetic
forces acting on the wire. What is the direction of the magnetic field?
F 0 
sin   0    0, or   180
Example: A horizontal wire carries a current and is in a vertical magnetic field.
What is the direction of the force on the wire?
1) left
2) right
I
3) zero
4) into the page
B
5) out of the page
Example: A vertical wire carries a current and is in a vertical magnetic field.
What is the direction of the force on the wire?
1) left
I
2) right
3) zero
4) into the page
B
5) out of the page
6. Current loop in magnetic field
Example: A rectangular current loop is in a uniform magnetic field.
What is the direction of the net force on the loop?
1) + x
Using the right-hand rule, we find that
2) + y
each of the four wire segments will
3) zero
experience a force outwards from the
4) - x
center of the loop. Thus, the forces
5) - y
of the opposing segments cancel, so
the net force is zero.
z
x
B
y
Wire and Loop with Electric Current in Magnetic Field
Battery
B
I
Axis of rotation
Fm
N
B
S
I
Fm
Fm
I
Magnetic field
Loop
The loop will rotate (clockwise)
S
N
Fm
N
I
S
N
B
S
Example: If there is a current in the loop in the direction shown, the loop will:
1) move up
F
2) move down
3) rotate clockwise
S
N
4) rotate counterclockwise
N
B
S
5) both rotate and move
B field out of North
B field into South
F
At the North Pole, the magnetic field points to the right and the current points out
of the page. The right-hand rule says that the force must point up. At the south
pole, the same logic leads to a downward force. Thus the loop rotates clockwise.
6a. Current loop and magnetic moment
B
Fa  IaB
Fb '
Fb  IbB sin  '



Fa   Fa '


F  0
Fb   Fb '
I
a
ab  A
Fa
I
Fb
M  IA
Fa '
b
  2 Fa sin   IaBb sin   IAB sin 
2
b

'
Fa 
B
  MB sin 
 N  N 1
M  NIA
N – number of loops in the coil
M - magnetic dipole moment:

 - angle between B and the perpendicu lar to the coil

 angle between Fa and the direction of the current in b
  MB sin 
 2

 2


Example: A current-caring circular coil is in uniform magnetic field. The
maximum magnitude of the torque on the coil occurs when the angle between
the field and magnetic moment is
0
45°
90°
135°
180°
Example: A coil is oriented initially with its magnetic dipole moment
antiparallel to a uniform magnetic field. In this orientation torque  is equal to:
A)  = 0
B)  = max
C) 0<  < max
Example:
N  10
A  r 2
r  0.10m
  30 
I  3.00 A
B  2.00T
M ?
 ?
M  NIA
M  10  3.00 A  0.10m2  0.94 Am2
  MB sin 
  0.94 A  m2  2.00T sin 30  0.94 N  m