2.2 ELECTROMAGNETISM
th
19
November 2012
Magnetic Fields
• A magnetic field exists around a moving charge in addition
to its electric field. A current carrying conductor produces
a circular field as shown below.
•The direction of the field is
described by the left Handed
Screw Rule, providing that we are
dealing with electron flow current.
If the thumb of the left hand points in the
direction of current (electron flow) then the
fingers show the field direction.
Magnetic Induction
• The strength of a magnetic field is called the magnetic induction,
B (or magnetic field density, or B-filed).
• It is measured in Tesla, T. The Tesla is defined as follows:
– “One Tesla is the magnetic induction in which a conductor of length
one metre, carrying a current of one Ampere, perpendicular to the
field, is acted on by a force of one Newton”
• A charged particle moving across a magnetic field experiences a
force. The magnitude of the force depends on the magnetic
induction, B, the current flowing (in the case of a current carrying
conductor), I, and the length of the conductor perpendicular to
the field.
• If the conductor lies perpendicular to the field then the force is
given by:
F = BIL
Table 4.2: Typical magnetic field values
Situation
Magnetic field (T)
Magnetic field of the Earth
5 x 10-5
At the poles of a typical fridge magnet
1 x 10-3
Between the poles of a large electromagnet
1.00
In the interior of an atom
10.0
Largest steady field produced in a laboratory
45.0
At the surface of a neutron star (estimated)
1.0 x 108
• In the general case where the
conductor lies at an angle to the B-Field
as shown:
L
I
Θ
LsinΘ
B
Θ
LcosΘ
F = BILsin Θ
LsinΘ
Direction of force
• The direction of the force is given by the right hand rule.
First Finger = Magnetic Field
from North to South
SeCond Finger =
CURRENT
(electron flow)
Thumb = force / motion
Example
Calculate the force on a wire of length 20cm at an angle of
30o to
a magnetic induction of 12T, if it is carrying a current of 3A.
L = 0.2m
Θ=
30o
B = 12T
F = 3A
F = BILsinΘ
= 20 x 3 x 0.2 x 0.5
= 3.6N
Measuring Magnetic Induction
• A known length of wire is placed perpendicular to the magnetic
induction between two permanent magnets on a sensitive
balance.
•When a current is passed
through the wire it
experiences a force.
•Similarly, the wire exerts a
force on the magnetic field
which causes the reading on
the balance to change.
•If the direction of the current is reversed then the sign of the reading,
(Δm) also reverses. We can convert Δm into force, and by measuring the
current, calculate B using:
B= F
IL
The direction of B can be
found from the right hand
rule.
Force Between Current Carrying Wires
• The magnetic induction around a current carrying
wire has the shape shown below.
.
Magnetic field due to a
electron current travelling
out of the paper.
x
Magnetic field due to a electron
current travelling into the paper.
Remember that the direction of the magnetic field is obtained
using the left handed screw rule for (electron flow) current.
• In general, the magnetic induction, B, at a distance, r,
from an infinite straight conductor carrying a current, I,
is given by:
0 I
B
2r
Where, μ0 is the permeability of
free space, 4π x10-7 TmA-1
Example
Calculate the magnetic induction 220cm from a long straight
wire carrying a current of 3A
•B = ?
•I = 3A
•r = 2.2m
0 I
B
2r
-7 3
4


10

2 2.2
 2.73107T
• Consider two parallel wires of infinite length separated by a
distance r and carrying currents I1 and I2 in the same
direction.
If the wires are separated by a distance r,
the magnetic induction at wire 2 due to the
current in wire 1 is:
 0 I1
B1 
2 r
So wire 2, carrying a current I2 , will
experience a force (along length, L)
F2 = B1I2L
Substitute
for B1
F
L
 0 I1
F2 
 I 2 L Similarly, wire 1
experiences a
2 r
F μ0 I1 I 2

L
2r
is known as the force per unit length
This is contents statement 8 DERIVE
force due to the
magnetic
induction
around wire 2
Note:
• For wires carrying current in the same direction,
the forces are attractive.
• For wires carrying currents in opposite directions,
the forces are repulsive.
Example
Two long parallel wires are 5cm apart. They exert a force
per unit length of 6x10-7Nm-1 on each other. If one wire
carries a current of 400mA, calculate the current in the
second wire.
F μ0 I1 I 2
r = 0.05m

L
I1 = 0.4A
 2 107 Nm1
I2= ?
F/L = 6x10-7Nm-1
F
L
2 r
I 2  375mA
Is known as the force per unit length
Definition of the Ampere
• A current of one ampere is defined as the constant current which,
if in two straight parallel conductors of infinite length placed one
metre apart in a vacuum, will produce a force between the
conductors of 2 x 10-7 Newton's per metre.
• To confirm this definition apply F  μ0 I1 I 2
L
2r
• to this situation.
Thus I1 and I2 both
equal 1 A, r is 1 m and
μo = 4π x 10-7 N A-2.
F 4 107 11

L
2 1
7
 2 10 Nm
1