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.

B

 1

r

• Thus we combine these together to get

B



I r

•as an equation

B



kI r

•for mathematical & historical reasons the constant k has been written as

k

  2  0

• however we shall for simplicity use "k“ • Our formula then is

B



kI 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?

B



B



kI

2

r x

10  7 0 .

15 ( 25 )

B

 3 .

33

x

10  5 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, 

kI B



N r

 2 4

x

10  4  12

x

10 0 .

12  7

I 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