Magnetism and Magnetic Effects

Electrons spinning on their axes are magnetic dipoles that may or may not cancel out each other depending on the type of dipole.
We measure the  magnetic moment of electron in Joule / Tesla.
The value of magnetic dipole of an electron is $9.27 \times 10^{-24}$ Joule/Tesla.   

According to Bohr modal of hydrogen like atoms, negatively charged electrons revolves around the positively charged nucleus. This uniform circular motion of electrons is equivalent to a current loop which possesses a magnetic dipole moment =IA

I:- current in the loop

A:-area

Consider an electron revolving anticlockwise around a nucleus in an orbit of radius r with speed v and time period T.

Equivalent current,

I= charge/time

=e/t = e/2πr/v

=ev/2πr

Area of current loop ,A=πr²

So , orbital magnetic moment of electron is=IA

=(ev/2πr). πr²

=evr/2

As J is angular momentum. J=mvr

m- mass of electron

So orbital magnetic moment = eJ/2m

Kinetic energy = qV where q is charge and V is potential
But potential is same
So, $K. E \propto q$
$q _p = q _e = 2q$
Putting these values,
we get the ratio of 1:1 :2

The radius of a charged particle in a magnetic field is r = p / (qB). Since both particles have the same momentum p and enter the same field B, the ratio r1/r2 = q2/q1. A proton has charge 1 and an alpha particle has charge 2, so r1/r2 = 2/1 = 2.

It velocity of particle is perpendicular to the magnetic field,the particles moves in a circle.The centripetal force is experienced by electron is $qVB$

When a charged particle is projected at an angle $\theta$ to a magnetic field, the component of velocity parallel to the field is $v\cos \theta$ while perpendicular to the field is $v\sin \theta$, so the particle will move in a circle of radius
$r = \dfrac{m(v\sin \theta)}{qB}=\dfrac{( 1.67 \times 10^{-27}) \times \left( 4 \times 10^{5} \times \sin 60^0 \right)}{1.6 \times 10^{-19} \times 0.3}=\dfrac{ \left( 1.67 \times 10^{-27} \right) \times \left( 4 \times 10^{5} \times \dfrac{\sqrt{3}}{2} \right)}{1.6 \times 10^{-19} \times 0.3}=\dfrac{2 \times 10^{-2}}{\sqrt{3}}$
Time period: $T = \dfrac{2\pi r}{vsin\theta}= \dfrac{2\pi \times \dfrac{2 \times 10^{-2}}{\sqrt{3}}}{4 \times 10^5 \times \sin 60^0}= \dfrac{2\pi \times \dfrac{2 \times 10^{-2}}{\sqrt{3}}}{4 \times 10^5 \times \dfrac{\sqrt{3}}{2}}=\dfrac{2\pi}{3} \times 10^{-7}$
Pitch: $P = v \cos \theta T= (4\times 10^5) \times \cos 60^0 \times \dfrac{2\pi}{3} \times 10^{-7}=\dfrac{4\pi}{3}\times 10^{-2}= 4.35 \times 10^{-2}: m$

Given that,

The radius of particle X$={{R} _{1}}$

The radius of particle Y$={{R} _{2}}$

We know that,

Work done of each particle = $qV$

Now, suppose the particle starts from rest, and final kinetic energy is

For, X particle,

$\dfrac{1}{2}{{m} _{1}}v _{1}^{2}=qV$

For, Y particle

$\dfrac{1}{2}{{m} _{2}}v _{2}^{2}=qV$

Now,

${{m} _{1}}v _{1}^{2}={{m} _{2}}v _{2}^{2}.....(I)$

Now, from the magnetic force is

$ {{F} _{{{m} _{1}}}}=q{{v} _{1}}B $

$ {{F} _{{{m} _{2}}}}=q{{v} _{2}}B $

Now, the magnetic force is equal to the centripetal force is

For, X

$ \dfrac{{{m} _{1}}v _{1}^{2}}{{{R} _{1}}}=qB{{v} _{1}} $

$ {{m} _{1}}{{v} _{1}}=qB{{R} _{1}} $

For, Y

$ \dfrac{{{m} _{2}}v _{2}^{2}}{{{R} _{2}}}=qB{{v} _{2}} $

$ {{m} _{2}}{{v} _{2}}=qB{{R} _{2}} $

Now, putting the value in equation (I)

$ {{m} _{1}}{{\left( \dfrac{qB{{R} _{1}}}{{{m} _{1}}} \right)}^{2}}={{m} _{2}}{{\left( \dfrac{qB{{R} _{2}}}{{{m} _{2}}} \right)}^{2}} $

$ \dfrac{R _{1}^{2}}{{{m} _{1}}}=\dfrac{R _{2}^{2}}{{{m} _{2}}} $

$ \dfrac{{{m} _{1}}}{{{m} _{2}}}={{\left( \dfrac{{{R} _{1}}}{{{R} _{2}}} \right)}^{2}} $

 Hence, the ratio of the mass is ${{\left( \dfrac{{{R} _{1}}}{{{R} _{2}}} \right)}^{2}}$

The electron moves in a helical path in a uniform magnetic field. The radius of the helix is r = (mv*sin(alpha))/(eB) and the pitch angle/period determines the displacement. The distance from the axis is calculated using the geometry of the circular projection of the helical motion.

We know$,$

Diamagnetic materials have no permanent magnetic moment; the induced magnetic moment opposes the external magnetic field, resulting in a net magnetic moment of zero in the absence of an external field.

For equilibrium of dipole net torque on the loop should be zero.
So, $\vec{Z} = \vec{M}\times \vec{B}$
$\vec{Z} _1 = \vec{Z} _o$
$\vec{M}\times \vec{B} = \vec{M}\times \vec{B} _o$
$MB \sin \theta = M B _o \sin (90^o-\theta)$
$B \sin\theta = B _o \cos \theta$   $(\because \sin(90^o-\theta) = \cos \theta)$

Torque $(\tau) = \vec{M}\times \vec{B}$
                   $= \vec{P _m}\times \vec{B}$