Magnetism Concept Page - 11

Example

Current carrying conductors in magnetic field involving Newton's Laws

A conducting rod of length l$l$ and mass m$m$ is moving down a smooth inclined plane of inclination θ$\theta$ with constant velocity V$V$. A current i$i$ is flowing in the conductor in a direction perpendicular to paper inwards. The magnitude of field B⃗ $\overrightarrow{B}$ is (Given acceleration due to gravity =g$=g$)F=Mgsinθ−Bilcosθ$F=Mgsin\theta -Bilcos\theta$
Since particle is moving down with constant magnetic field
F=0$F=0$
mgsinθ−Bilcosθ=0$mgsin\theta -Bilcos\theta =0$
B=mgiltanθ$B={\displaystyle \frac{mg}{il}}tan\theta$

Example

Potential energy of a magnet in a magnetic field

If a bar magnet in magnetic moment m$m$ is deflected from an angle θ$\theta$ in a uniform magnetic field of induction B$B$, the energy stored in reversing the direction is :Now lets assume that the bar magnetic is initially placed parallel to the field the initial potential energy will be Ui=−M⃗ .B⃗ =−MBcos0=−MB$U_i=-\overrightarrow{M}.\overrightarrow{B}=-MB\cos 0=-MB$
Now when it is deflected by an angle θ$\theta$ then Uf=−M⃗ .B⃗ =−MBcosθ$U_f=-\overrightarrow{M}.\overrightarrow{B}=-MB\cos \theta$
ΔU=Ui−Uf$\mathrm{\Delta }U=U_i-U_f$
ΔU=−MB(1−cosθ)$\mathrm{\Delta }U=-MB(1-\cos \theta )$

Example

Potential energy of a current loop in a magnetic field

A planar coil of area 7$7$ m2${\mathrm{m}}^2$ carrying an anti-clockwise current 2 A$2A$ is placed in an extemal magnetic field B⃗ =(0.2i^+0.2j^−0.3k^)$\overrightarrow{B}=(0.2\hat{i}+0.2\hat{j}-0.3\hat{k})$, such that the normal to the plane is along the line(3i^−5j^+4k^)$(3\hat{i}-5\hat{j}+4\hat{k})$Area A=7m2,$A=7m^2,$ Normal n^=150−−√(3i^−5j^+4k^)$\hat{n}={\displaystyle \frac{1}{\sqrt{50}}}(3\hat{i}-5\hat{j}+4\hat{k})$
≃17(3i^−5j^+4k^)$\simeq {\displaystyle \frac{1}{7}}(3\hat{i}-5\hat{j}+4\hat{k})$
∴$\therefore$ area A⃗ =(3i^−5j^+4k^)m2$\overrightarrow{A}=(3\hat{i}-5\hat{j}+4\hat{k})m^2$
Magnetic field B⃗ =(0.2i^+0.2j^−0.3k^)T$\overrightarrow{B}=(0.2\hat{i}+0.2\hat{j}-0.3\hat{k})T$
Current I=2 A$I=2A$
∴$\therefore$ magnetic moment M⃗ =IA⃗ =(6i^−10j^+8k^)Am2$\overrightarrow{M}=I\overrightarrow{A}=(6\hat{i}-10\hat{j}+8\hat{k})A{m}^2$
|M⃗ |≃14 Am2(2×7=14)$|\overrightarrow{M}|\simeq 14A{m}^2(2\times 7=14)$
Potential energy U=−M⃗ .B⃗ $=-\overrightarrow{M}.\overrightarrow{B}$
=−[1.2−2−2.4]=3.2J$=-[1.2-2-2.4]=3.2J$

Example

Magnetic dipole moment of a revolving electron

An electron revolving in an orbit of radius 0.5 A˚$\mathring{A}$ in a hydrogen atom executes 10 revolutions per second. Find the magnetic moment of electron due to its orbital motion.
Radius of orbit =0.50A$={0.5}^{0}A$
I=ChargeTime=10×1.6×10−191=1.6×10−18$I={\displaystyle \frac{Charge}{Time}}={\displaystyle \frac{10\times 1.6\times {10}^{-19}}{1}}=1.6\times {10}^{-18}$

Magnetic moment =Iπr2$=I\pi r^2$

1.6×10−18×π×(0.5×10−10)2$1.6\times {10}^{-18}\times \pi \times (0.5\times {10}^{-10}{)}^2$

1256.63×10−(38+3)=1256×10−41Am2$1256.63\times {10}^{-(38+3)}=1256\times {10}^{-41}A{m}^2$

Definition

Gauss Law

In physics, Gauss's law for magnetism is one of the four maxwell equations that underlie classical electrodynamics. It states that the magnetic field B has divergence equal to zero, in other words, that it is a solenoidal vector field. It is equivalent to the statement that magnetic monopoles do not exist. Rather than "magnetic charges", the basic entity for magnetism is the magnetic dipole. 

Example

Magnetic dipole moment of a current loop in a magnetic field

m=niA$m=niA$
 where, m$m$ is the magnetic moment.

A circular current carrying coil has a radius r$r$. Its magnetic dipole moment is proportional to:Magnetic moment =niA$=niA$
                             =niπr2$=ni\pi r^2$      (∵A=πr2)$(\because A=\pi r^2)$

Definition

Gyromagnetic ratio

The gyromagnetic ratio (also sometimes known as the magnetogyric ratio in other disciplines) of a particle or system is the ratio of its magnetic dipole moment to its angular momentum, and it is often denoted by the symbol γ$\gamma$. Its SI unit is the radian per second per tesla (rads−1T−1$rad{s}^{-1}{T}^{-1}$) or, equivalently, the coulomb per kilogram (Ckg−1$(Ck{g}^{-1}$).

Definition

Orbital and spin magnetic moment

Any charge in uniform circular motion would have an associated magnetic moment labelled as the orbital magnetic moment.Besides the orbital moment, the electron has an intrinsic magnetic moment, called the spin magnetic moment.

Definition

Comparison of magnetic fields in axial position and equatorial position

For small magnet, the intensity of field on the axis is given by
B1=μo2M4πd3$B_1={\displaystyle \frac{{\mu }_{o}2M}{4\pi d^3}}$.................(1)
Intensity of field on neutral axis for the same distance is 
B2=μoM4πd3$B_2={\displaystyle \frac{{\mu }_{o}M}{4\pi d^3}}$...................(2)
Hence by Eq (1) and (2) we get, 
B1=2B2$B_1=2B_2$
For the same distance, the field on the axis is double of that on the neutral axis.

Example

Magnetic dipole moment of a rotating disc with given surface charge

A flat circular disc of radius R has uniform charge distribution of -σ$\sigma$ upto r$\mathrm{r}$ and of +σ$+\sigma$ from r$\mathrm{r}$ upto R$\mathrm{R}$, such that net charge on the disc is zero. lf the disc is rotating uniformly at the rate of ω$\omega$ rads−1${\mathrm{s}}^{-1}$, its magnetic moment then is given by πσωR4n${\displaystyle \frac{\pi \sigma \omega {\mathrm{R}}^4}{\mathrm{n}}}$. Find the value of n$n$.Given −σπr2+σπ(R2−r2)=0$-\sigma \pi r^2+\sigma \pi (R^2-r^2)=0$
∴r=R2√$\therefore r={\displaystyle \frac{R}{\sqrt{2}}}$
consider a circular strip of radius r$r$,width dr$dr$.
Magnetic moment dM=2πrdrσ′.πr2.ω2π$=2\pi rdr{\sigma }^{\prime }.\pi r^2.{\displaystyle \frac{\omega }{2\pi }}$
σ′=−σ,0≤r≤R2√${\sigma }^{\prime }=-\sigma ,0\le r\le {\displaystyle \frac{R}{\sqrt{2}}}$
=+σ,R2√≤r≤R$=+\sigma ,{\displaystyle \frac{R}{\sqrt{2}}}\le r\le R$
∴M=∫R0πωσ′r3dr=πωσ⎡⎣⎢⎢⎢−∫R2√0r3dr+∫RR2√r3dr⎤⎦⎥⎥⎥$\therefore M={\int }_0^R\pi \omega {\sigma }^{\prime }{r}^{3}dr=\pi \omega \sigma [-{\int }_0^{{\displaystyle \frac{R}{\sqrt{2}}}}{r}^{3}dr+{\int }_{{\displaystyle \frac{R}{\sqrt{2}}}}^R{r}^{3}dr]$
=πωσ[−R44×4+R44−R44×4]=πωσR4[−18+14]$=\pi \omega \sigma [-{\displaystyle \frac{R^4}{4\times 4}}+{\displaystyle \frac{R^4}{4}}-{\displaystyle \frac{R^4}{4\times 4}}]=\pi \omega \sigma R^4[-{\displaystyle \frac{1}{8}}+{\displaystyle \frac{1}{4}}]$
∴M=πωσR48=πωσR4n⇒n=8

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