Magnetic Field Strength Around
a Wire
From the demonstration, we saw
that:
• the magnetic field strength varies directly with the
amount of current flowing through the wire
• i.e. B  I
• the magnetic field strength varies inversely with
the distance from the wire.
• i.e.
1
B
r
• Thus we combine these together to get
I
B
r
•as an equation
kI
B
r
•for mathematical & historical reasons
the constant k has been written as
0
k
2
• however we shall for simplicity use "k“
• Our formula then is
kI
B
r
•where B = magnetic field strength (Tesla)
•I = current (Amps)
•r = distance from wire (m)
•k = permeability of free space constant = 2 x 10 -7
• Sample: A vertical wire carries a current of
25.0 A. What is the magnetic field strength
15 cm from the wire?
kI
B
r
7
2 x10 (25)
B
0.15
5
B  3.33 x10 T
Magnetic Field Strength inside a
Loop
• We predict the field strength will be greater
inside a loop. There are wires all around the
region exerting a magnetic field. The lines
of flux reinforce each other. The formula for
a single loop is
B
kI
r
r = radius of coil
• For a coil, the strength depends on the
number of coils. (N)
BN
kI
r
• Sample: If the magnetic field strength at the centre of a
loop of 12 coils is 4.00 x 10-4 T, and the radius is 12.0 cm,
find the current that is flowing,
BN
4 x10
4
kI
r
7
 2 x10 I
 12
0.12
I  6.37 A
Magnetic Field Strength inside a
Solenoid
• The formula for magnetic field strength inside a
solenoid is
• B = 2knI
where n = number of turns / m
Sample: A 10 cm long solenoid has 400 turns of wire
and carries a current of 2.00 A. Calculate the
magnetic field strength inside the solenoid.
n = 400 turns / 0.1m
= 4000 / m
B = 2knI
B = 22 x 10-7(4000)(2)
B=1.0 x 10-2 T