Magnetism Concept Page - 12

Example

Magnetic dipole moment of a rotating sphere with given volume charge

A charge q$q$ is uniformly distributed over the volume of a uniform ball of mass m$m$ and radius R$R$ which rotates with an angular velocity ω$\omega$ about the axis passing through its centre.Find the respective magnetic moment.Here, a charge q$q$ is uniformly distributed over the volume of a uniform ball of mass m$m$ and radius R$R$ which rotates with an angular velocity ω$\omega$ about the axis passing through its center.Find the volume of the ball.

V=43πR3$V={\displaystyle \frac{4}{3}}\pi R^3$
The charge dq$dq$ contained by the volume dV$dV$ of the ball isdq=q43πR3dV$dq={\displaystyle \frac{q}{{\displaystyle \frac{4}{3}}\pi R^3}}dV$................................(1)
Also, current dI$dI$ constituted by the same element is
dI=dqdt$dI={\displaystyle \frac{dq}{dt}}$

dI=3q4πR3dVdt$dI={\displaystyle \frac{3q}{4\pi R^3}}dVdt$
dI=3q4πR3dVω2π$dI={\displaystyle \frac{3q}{4\pi R^3}}dV{\displaystyle \frac{\omega }{2\pi }}$............($($since, ω=2πn$\omega =2\pi n$ and dt=1n)$dt={\displaystyle \frac{1}{n}})$
The magnetic moment along the axis of rotation in a circular arc is
Pm=∫R0dIdS$P_m={\int }_0^{R}dIdS$Pm=∫R03q4πR3dVω2π.πr2sin2θdr$P_m={\int }_0^R{\displaystyle \frac{3q}{4\pi R^3}}dV{\displaystyle \frac{\omega }{2\pi }}.\pi r^2{\sin }^2\theta dr$ 

Pm=∫R03q4πR3dVω2.r2sin2θdr$P_m={\int }_0^R{\displaystyle \frac{3q}{4\pi R^3}}dV{\displaystyle \frac{\omega }{2}}.r^2{\sin }^2\theta dr$ 
Now, integrating for whole volume of ball we get

Pm=∫π20∫R03q4πR3(2πr2sinθdθ)ω2.r2sin2θdr$P_m={\int }_0^{{\displaystyle \frac{\pi }{2}}}{\int }_0^R{\displaystyle \frac{3q}{4\pi R^3}}(2\pi r^2\sin \theta d\theta ){\displaystyle \frac{\omega }{2}}.r^2{\sin }^2\theta dr$ 

Pm=3ωq4R3∫π20∫R0r4.sin3θdθdr$P_m={\displaystyle \frac{3\omega q}{4R^3}}{\int }_0^{{\displaystyle \frac{\pi }{2}}}{\int }_0^R{r}^4.{\sin }^3\theta d\theta dr$

Pm=(3ωq4R3)(4R55)(13)$P_m=\left({\displaystyle \frac{3\omega q}{4R^3}}\right)\left({\displaystyle \frac{4R^5}{5}}\right)\left({\displaystyle \frac{1}{3}}\right)$

Pm=qR2ω5$P_m={\displaystyle \frac{q{R}^2\omega }{5}}$

Definition

Circular current loop as a magnetic dipole

Magnetic field lines due to a current carrying wire are shown in the attached plot. The field lines resemble that of a magnet. Hence, it behaves as a magnetic dipole. Its poles can be found from its field lines (Field lines leave the magnet from the north pole and enter from the south pole).

Definition

Equivalence between current carrying circular coil and magnetic dipole

Magnetic dipole moment M$M$ with the circular current loop carrying a current I$I$ and of area A$A$.  The magnitude of m$m$ is given by 
|m|=IA$|m|=IA$
Further, the direction of the magnetic dipole moment is perpendicular to the plane of the loop.

Example

Torque on a bar magnet placed in an uniform magnetic field

Couple:$Couple:$
We can define couple as a pair of two equal and opposite forces having different lines of action. They give rise to turning effect known as torque along the axis which is perpendicular to the plane of forces. 
   Torque=Magnetic force×Perpendicular distance$Torque=Magneticforce\times Perpendiculardistance$
   τ=mB×2lsinθ$\tau =mB\times 2lsin\theta$
   τ=m(2l)Bsinθ$\tau =m(2l)Bsin\theta$
   τ=M⃗ ×B⃗ $\tau =\overrightarrow{M}\times \overrightarrow{B}$
where M = Magnetic moment 
           B = Magnetic field
A bar magnet of magnetic moment 1.5 J/T is aligned with the direction of a uniform magnetic field of 0.22 T. Find work done in turning the magnet so as to align its magnetic moment opposite to the field and the torque acting on it in this position.M=1.5 J/T$M=1.5J/T$
B=0.22 T$B=0.22T$
Magnetic moment in an externally produced magnetic field has potential energy as:
U=−M⃗ .B⃗ $U=-\overrightarrow{M}.\overrightarrow{B}$
U1=−1.5×2.210=−0.33T$U_1=-\frac{1.5\times 2.2}{10}=-0.33T$
U2=1.5×2.210=0.33T$U_2=\frac{1.5\times 2.2}{10}=0.33T$
workdone=ΔU$workdone=\mathrm{\Delta }U$
U2−U1$U_2-U_1$
=2U$=2U$
=0.66J$=0.66J$
τ⃗ =M⃗ ×B⃗ $\overrightarrow{\tau }=\overrightarrow{M}\times \overrightarrow{B}$
=MBsinθ$=MBsin\theta$
=0(∵θ=π)$=0(\because \theta =\pi )$

Definition

The magnitude of magnetic induction at a point along the axis for short magnetic dipole

If BP$B_P$ is the total magnitude of magnetic induction at a point along the axis of magnetic dipole then
BP=μ04π.m(π−d)2−μ04π.m(π−d)2$B_P={\displaystyle \frac{{\mu }_0}{4\pi }}.{\displaystyle \frac{m}{{(\pi -d)}^2}}-{\displaystyle \frac{{\mu }_0}{4\pi }}.{\displaystyle \frac{m}{{(\pi -d)}^2}}$
BP=μ0m4π.2μr(r2−d2)2$B_P={\displaystyle \frac{{\mu }_{0}m}{4\pi }}.{\displaystyle \frac{2\mu r}{{(r^2-d^2)}^2}}$ where μ=2md$\mu =2md$ which can be represented as magnetic dipole M$M$.

Definition

Induced magnetism

When a magnetic material (like iron) is brought close to a magnet, it acts like a temporary magnet and is attracted to it. This phenomenon is called induced magnetism. Induced magnetism is usually lost when the strong magnet is removed.

Definition

Magnetic induction

Magnetic induction or magnetic field is the magnetic effect of electric currents and magnetic materials. The magnetic field at any given point is specified by both a direction and a magnitude (or strength); as such it is a vector field.Magnetic fields can be produced by moving electric charges and the intrinsic magnetic moments of elementary particles associated with a fundamental quantum property, their spin.Often the magnetic field is defined by the force it exerts on a moving charged particle.

Definition

Intensity of magnetisation

A quantity used in describing magnetic phenomena in terms of their magnetic fields and magnetization. 
H=Bμ0−M$H={\displaystyle \frac{B}{{\mu }_0}}-M$
where H$H$ is magnetic intensity,
M$M$ is magnetisation and 
B$B$ is the magnetic field

Definition

Magnetic intensity or magnetic field strength

Consider a material placed in an external magnetic field B0$B_0$. The magnetic intensity is defined as,
H=B0μ0$H={\displaystyle \frac{B_0}{{\mu }_0}}$
where μ0${\mu }_0$ is the permeability of free space

Example

Magnetic permeability and relative magnetic permeability

μ=μ0μr$\mu ={\mu }_0{\mu }_r$
where, μ$\mu$ is the magnetic permeability and μr${\mu }_r$ is the relative magnetic permeability.
The absolute permeability of a medium is 8.792×10$8.792\times 10$ −6${}^{-6}$ henry/m. Its relative permeability is:

μr=μμ0=8.792×10−61.2566×10−6=7

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