# 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 $\overrightarrow{E}&space;=&space;\frac{\overrightarrow{F}}{Q}$  .

$\overrightarrow{F_{e}}&space;=&space;Q&space;\overrightarrow{E}-----EQN(1)$ .

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

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

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 $\overrightarrow{F_{m}}.\overrightarrow{dl}&space;=0$ 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 $\overrightarrow{F}&space;=&space;\overrightarrow{F_{e}}+\overrightarrow{F_{m}}$ .

$\overrightarrow{F}&space;=&space;Q(\overrightarrow{E}+\overrightarrow{u}X\overrightarrow{B})$ .

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 $\overrightarrow{F}&space;=&space;m\overrightarrow{a}$ .

$\overrightarrow{F}&space;=&space;m\frac{d\overrightarrow{u}}{dt}$ .

$\overrightarrow{F}&space;=&space;Q(\overrightarrow{u}X\overrightarrow{B})$ .

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

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

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 $\overrightarrow{B}$ .

$I\overrightarrow{dl}\approx&space;\overrightarrow{k}ds\approx&space;\overrightarrow{J}dv$ .

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

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

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

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

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

$d\overrightarrow{F}=dQ\overrightarrow{u}X\overrightarrow{B}$.

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

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

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

i.e,

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

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

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)    $\overrightarrow{F_{m}}&space;=&space;Q\overrightarrow{u}&space;X&space;\overrightarrow{B}\Rightarrow&space;\frac{\overrightarrow{F_{m}}}{Q}=\overrightarrow{u}&space;X&space;\overrightarrow{B}$    similar to $\overrightarrow{E}&space;=&space;\frac{\overrightarrow{F_{e}}}{Q}$ .

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 s$I_{1}\overrightarrow{dl_{1}}$ 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  $d\overrightarrow{B_{2}}$ .

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

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

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

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

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

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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