Calculating Magnetic Fields – Learn


The strength of the magnetic field around a current-carrying conductor and a solenoid can be determined quantitatively.

  • The equation used to calculate the strength of the field around a current-carrying conductor, gives us the magnitude of the field at some distance, r, from the conductor.
  • The equation used to calculate the strength of the field in a solenoid, gives us the magnitude of the field at a point inside of the solenoid. The field within the solenoid is uniform.

Current-carrying conductors

The magnitude of the field (\vec { B } ) at some distance, r, from the current-carrying conductor is given by:

\vec { B } =\cfrac { { \mu }_{ 0 }I }{ 2\pi r }

where:

\vec { B } = the strength of the magnetic field (in Tesla)

 { \mu }_{ 0 }  = the permeability of free space ({ 1.257\times 10 }^{ -6 }N{ A }^{ -2 } )

I = the current in the wire (in amps)

r = the radius, or distance from the wire (in m)


Solenoids

The magnitude of the field (\vec { B } ) within a solenoid is given by:

\vec { B } =\cfrac { { \mu }_{ 0 }NI }{ L }

where:

\vec { B } = the strength of the magnetic field (in Tesla)

 { \mu }_{ 0 }  = the permeability of free space ({ 1.257\times 10 }^{ -6 }N{ A }^{ -2 } )

N = the number of turns in the coil

I = the current in the wire (in amps)

L = the length of the solenoid (in m)


Example 1:

What is the strength of a magnetic field at a distance of 5cm from a wire carrying a current of 10A?

Answer:

\vec { B } =\cfrac { { \mu }_{ 0 }I }{ 2\pi r }

where:

 { \mu }_{ 0 }  =  ({ 1.257\times 10 }^{ -6 }N{ A }^{ -2 } )

I\:= \:10 \:amps

r\:=\:0.05m

\vec { B } =\cfrac { { { 1.257\times 10 }^{ -6 }\: }\times \: 10 }{ 2\pi \:\times\: 0.05 }

\vec { B } =\:{ 4\times 10 }^{ -5 }


Example 2:

What is the strength of a magnetic field in a 15cm solenoid that has 250 turns carrying a current of 20A?

Answer:

\vec { B } =\cfrac { { \mu }_{ 0 }NI }{ L }

where:

 { \mu }_{ 0 }  =  ({ 1.257\times 10 }^{ -6 }N{ A }^{ -2 } )

I\:= \:20 \:amps

L\:=\:0.15m

N\:=\:250

\vec { B } =\cfrac { { 1.257\times 10 }^{ -6 }\: \times \: 250\: \times \:20 }{ 0.15 }

\vec { B } = 0.0419\:T

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