Calculating Magnetic Fields - Learn - ScienceFlip

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:

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

$I\:= \:20 \:amps$

$L\:=\:0.15m$

$N\:=\:250$

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

$\vec { B } = 0.0419\:T$