Magnetic Forces

Magnetic forces are required to study the force , a magnetic field exerts on charged particles, current elements and loops which is used in electrical devices in ammeters, volt meters, Galvano meters.

There are 3 ways in which force due to magnetic fields can be experienced.

  1. The force can be due to a  moving charged particle in a  Magnetic field.
  2. on a current element in an external B  field.
  3. between two current elements.

Force on a charged particle:-

we know that \overrightarrow{E} = \frac{\overrightarrow{F}}{Q}  .

\overrightarrow{F_{e}} = Q \overrightarrow{E}-----EQN(1) .

where  \overrightarrow{F_{e}}  is the electric force on a stationary (or) moving electric charge in an electric field and is related to  \overrightarrow{E} .  where  \overrightarrow{F_{e}}  and   \overrightarrow{E} are in the same direction.

a magnetic field can exert force only on a moving charge , suppose a charge Q is moving with velocity u  (or) v in a magnetic field (B) is

\overrightarrow{F_{m}} = (Q\ \overrightarrow{u} \ X \overrightarrow{B})-----EQN(2) .

from the  equations    \overrightarrow{F_{e}}   is independent of velocity of the charge and performs work on the charge which changes its kinetic energy but  \overrightarrow{F_{m}}  depends on the charge velocity and is normal to it so work done \overrightarrow{F_{m}}.\overrightarrow{dl} =0 it does not cause increase in the kinetic energy of the charge.

 \overrightarrow{F_{m}}  is small compared to    \overrightarrow{F_{e}}  except at high velocities.

so a charge which is in movement has both electric and magnetic fields.

Then \overrightarrow{F} = \overrightarrow{F_{e}}+\overrightarrow{F_{m}} .

\overrightarrow{F} = Q(\overrightarrow{E}+\overrightarrow{u}X\overrightarrow{B}) .

This is known as Lorentz’s force equation. It relates mechanical force to electrical force.

if the mass of the charges particle is m

Then \overrightarrow{F} = m\overrightarrow{a} .

\overrightarrow{F} = m\frac{d\overrightarrow{u}}{dt} .

\overrightarrow{F} = Q(\overrightarrow{u}X\overrightarrow{B}) .

m\frac{d\overrightarrow{u}}{dt}= Q(\overrightarrow{u}X\overrightarrow{B}) .

\frac{d\overrightarrow{u}}{dt}-\frac{1}{m} (\overrightarrow{u}X\overrightarrow{B})=Qm\overrightarrow{E} .

The solution to this equation is  important in determining the motion of charged particles in in such cases the energy transfer is only by means of electric field.

Force on a current element:-

Consider a current carrying conductor I\overrightarrow{dl} , in order to find out the force acting on the current carrying element by the  magnetic field \overrightarrow{B} .

I\overrightarrow{dl}\approx \overrightarrow{k}ds\approx \overrightarrow{J}dv .

I\overrightarrow{dl}=\frac{dQ}{dt}\overrightarrow{dl} .

I\overrightarrow{dl}=dQ\frac{\overrightarrow{dl}}{dt} .

I\overrightarrow{dl}=dQ\overrightarrow{u} .

I\overrightarrow{dl} is nothing but a elemental charge dQ moving with the velocity \overrightarrow{u} .

\overrightarrow{F}=Q\overrightarrow{u}X\overrightarrow{B} .


d\overrightarrow{F}=I\overrightarrow{dl}X\overrightarrow{B} .

\overrightarrow{F}=\oint_{l}I\overrightarrow{dl}X\overrightarrow{B} .

The line integral is for the current is along the closed path.


The magnetic field produced by the current element  I\overrightarrow{dl}  does not exert force on the element itself just as a point charge does not exert force on itself.

So the  magnetic field  \overrightarrow{B}  that exerts force on I\overrightarrow{dl} must be from the another element in other words the magnetic field  \overrightarrow{B}is external to the current element I\overrightarrow{dl}.

Similarly we have  force equations for other current elements  \overrightarrow{k}ds  and \overrightarrow{J}dv as follows

\overrightarrow{F}=\oint_{s}\overrightarrow{k} dsX\overrightarrow{B}       and        \overrightarrow{F}=\oint_{v}\overrightarrow{J} dvX\overrightarrow{B} .

So the magnetic field is defined as the force per unit current element

i.e,  \frac{d\overrightarrow{F}}{\overrightarrow{k}ds}=\overrightarrow{B}  .(or)    \overrightarrow{F_{m}} = Q\overrightarrow{u} X \overrightarrow{B}\Rightarrow \frac{\overrightarrow{F_{m}}}{Q}=\overrightarrow{u} X \overrightarrow{B}    similar to \overrightarrow{E} = \frac{\overrightarrow{F_{e}}}{Q} .

so the    \overrightarrow{B}  describes the force properties of a magnetic field.

Force between two current elements (Ampere’s force law):-

Consider two current loops  I_{1}\overrightarrow{dl_{1}}  and  I_{2}\overrightarrow{dl_{2}} then by Biot- Savart’s law  both current elements produces respective magnetic fields so we may find the force on element sI_{1}\overrightarrow{dl_{1}} due to the field produced by I_{2}\overrightarrow{dl_{2}}.

Field produced by current element  is   I_{2}\overrightarrow{dl_{2}}  is  d\overrightarrow{B_{2}} .

So the force applied on  the element   I_{1}\overrightarrow{dl_{1}} is    d\overrightarrow{F_{1}}  by the field  d\overrightarrow{B_{2}} .

d(d\overrightarrow{F_{1}}) = I_{1}\overrightarrow{dl_{1}}Xd\overrightarrow{B_{2}} ..

d\overrightarrow{B_{2}} = \frac{\mu _{o}I_{2}\overrightarrow{dl_{2}}X \widehat{{a_{R21}}}}{4\pi R_{21}^{2}} .

d(d\overrightarrow{F_{1}}) = \frac{\mu _{o}I_{1}\overrightarrow{dl_{1}}XI_{2}\overrightarrow{dl_{2}}X \widehat{{a_{R21}}}}{4\pi R_{21}^{2}}.

This is similar to coulomb’s law in electrostatics. Here it is law of force between two current elements and is analogous to coulomb’s law

\overrightarrow{F_{1}} =\frac{\mu _{o}I_{1}I_{2}}{4\pi}\oint_{l_{1}} \oint_{l_{2}}\frac{\ \overrightarrow{dl_{1}}X\ \overrightarrow{dl_{2}}X \ \widehat{{a_{R21}}}}{ R_{21}^{2}} .

Then the force  \overrightarrow{F_{2}}  acting on loop 2 due to the field produced by the current element  t I_{1}\overrightarrow{dl_{1}}   is nothing but \overrightarrow{F_{2}}=-\overrightarrow{F_{1}}

Note:- This is nothing but the Ampere’s force law that is the force between two current carrying conductors is given by it.


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Author: vikramarka

Completed M.Tech in Digital Electronics and Communication Systems and currently working as a faculty.