Transcript Forces on Current Carrying Wires in Magnetic Fields

Chapter 19
Forces on Current Carrying Wires in Magnetic
Fields
Herriman High School - AP Physics 2
Section 19.1: Magnets
 A magnet consists of two poles.
 If we consider a bar magnet one end is called the north pole
and the other the south pole.
 In a compass the north pole of the magnet points toward the
geographic north pole of the earth and the south pole
of the magnet points toward the geographic south pole
of the earth.
 Like electric charges – magnetic material follow the property
that opposites attract and likes repel.
 The Geographic North pole of the earth is actually its
magnetic south pole.
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Section 19.3: Magnetic Fields
 Just like an electric charge develops a field around it – a
magnet also creates a field effect around itself.
 Conventionally Magnetic Field lines point outward from
the north pole and terminate inward at the south pole.
 A stationary charged particle is not affected by a stationary
magnetic field, however a charged particle moving
through a magnetic field experiences a force. The
magnitude of this force is defined as:
F  qvB sin 
F  qv  B
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q = charge of particle
v = velocity of the particle in m/s
B is the magnetic field strength in tesla (T)
θ = the angle between the magnetic field and
the motion of the particle.
The Direction of the Magnetic Force on
a Charged Particle.
 B = magnetic field strength is measured in Tesla or weber per
square meter, 1 T = 1Wb/m2, often however it is more
convenient to use the gauss (G) where 1 T = 104 G.
 From the Force equation, you can also see that the force exerted is
at its maximum when the particle is moving perpendicular to
the magnetic field and drops to zero when the particle moves
parallel to the magnetic field.
 Determination of the direction of the force is done by using the
right hand rule.
 Point the fingers of the right hand in the direction the particle is
moving.
 Point your palm in the direction of the magnetic field.
 Your thumb will now point in the direction of the force.
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Sample Problem:
 A proton moves with a speed of 1 x 105 m/s in the earth’s
magnetic field which has a value of 55μT at a particular
location. When the proton moves eastward, the magnetic
force is directed upward, and when it moves northward, no
magnetic force acts on it.
a) What is the direction of the magnetic field?
b) What is the strength of the magnetic field when the proton
moves eastward?
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Solution
 Use the right hand rule to get the direction of the magnetic
field – north or south. (North)
 Substitute to get the magnitude of the field:
F  qvB sin 
F  (1.6 x1019 C )(1x105 m / s)(55x106 T )(sin 90)  8.8x1019 N
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Section 19.4: Magnetic Force on a
Current Carrying Conductor
 Force on a current carrying wire
 If a moving charged particle experiences a force because of
a magnetic field then it makes sense that a current carrying
conductor would too.
 (Remember that current is the movement of charge in a
certain amount of time.)
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Magnetic force on a wire
 The force on a wire follows
the same right hand rule.
F  BIl sin 
F  Il  B
B = Magnetic Field
I = current in amperes
L = length of the wire
θ = angle between the field and the
wire.
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Physics 2
Sample Problem:
 A wire carries a current of 22  Magnitude:
A directed west to east. The
 F = BIL sin θ
magnetic field at this location
 F = (5 x 10-4 T)(22 A)(36 m)sin 90°
is directed south to north
= .0396 N
with a strength of 0.5 x 10-4  Direction
T. Find the magnitude and
 Fingers point east, palm north, so
direction of the magnetic
thumb is upward. Force is upward.
force on a 36 meter length of
wire.
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Physics 2
Section 19.5: Torque on a current
carrying loop
 If the wire is bent in the form of a loop how will the forces
act?
 The two opposing forces will produce a torque
 Note that the Force will be the greatest when B is parallel to
the plane of the loop!
 Why? Because when it is parallel to the rectangular loop
part of the loop will be parallel to the magnetic field but the
other part will be exactly perpendicular
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  NBIA sin 
The Formula
 τ=torque
 N is the number of turns or loops in the wire
 B is the Mag field
 A is the area of the loop
 θ is the angle between the Magnetic field and the line
perpendicular to the loop
 μ is called the magnetic moment of the loop.
  IAN
So torque can also equal
  B sin 
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Physics 2
Sample Problem
 A circular wire of radius 1 m is placed in a magnetic field of
0.5 T. The normal of the plane of the loop makes an angle of
30 degrees with the magnetic field and the current is 2 A
moving counter clockwise around the loop.
 Find the magnetic moment of the loop and the magnitude of
the torque at this instant.
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Solution:
 μ = IAN = (2 A)(3.14 m2)(1) = 6.28 A*m2
 T = μBsinθ = (6.28 A*m2)(0.5 T)(sin 30) = 1.57 N*m.
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Section 19.7: Magnetic Field of a Long
Straight Wire and Ampere’s Law
 Current carrying wires
actually produce their own
magnetic field.
 The magnetic field
surrounds the wire.
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Right hand rule for wires
 Your thumb heads in the
direction of the current (I)
 Your fingers wrap around
the wire in the direction of
the magnetic field
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Physics 2
The Formula
 B stands for the magnetic field strength
 r is the distance from the wire
 μ0=4πx10-7 Tm/A
 AKA the permeability of free space.
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0 I
B
2r
Finding the Magnetic field for more
than one wire
 If there are two wires you need to find the magnetic field for
each wire
 Find the direction of the magnetic field for each wire
 If the magnetic fields are in the same direction you may add
 If they are in opposite directions subtract
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Sample Problem
Problem:
 A long straight wire carries a current of 5 A. Calculate the
magnetic field of the wire 4 mm away.
Solution:
0 I (4x107 )(5)
B

 2.5 x104 T
2r
2 (.004)
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Section 19.8: Magnetic Force Between
Two Parallel Conductors
 µ0 – permeability of free space
 d – the distance separating the 2
wires.
 The forces are attractive if the
currents are in the same
direction
 Repulsive if in opposite
directions.
F  0 I1 I 2

l
2d
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Physics 2
Sample Problem
 Two wires, each having a weight
 Fgrav + Fmag = 0
per unit length of
 0 I1 I 2
1 x 10-4 N/m, are parallel with
 mg 
l 0
one directly above the other.
2d
Assume the wires carry currents
0 I 2
that are equal in magnitude and
l  mg
2d
opposite in direction. The wires
(2d )( mg / l )
are 0.1 meters apart and the
2
I 
sum of the magnetic and
0
gravitational force on the upper
wire is zero. Find the current in
4
(
2
)(

)(
0
.
1
m
)(
1
x
10
)
2
the wires, neglect Earth’s
I 
 7.07 A
7
4x10
magnetic field.
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Physics 2