Magnetism Concept Page - 6

Example

Magnetic field due to concentric current carrying straight wires

Example:
Find the magnetic field due to the configuration of wires at a distance r$r$ from the axis as shown in the figure. The current is uniformly distributed in the cross sectional areas of 0<r<a$0<r<a$ and a<r<b$a<r<b$.
Solution:
Amperian loop is a circular loop of radius r concentric to the wires.
For r≤a$r\le a$, 
     Current enclosed in the amperian loop is I1=2r2a2$I_1={\displaystyle \frac{2r^2}{a^2}}$
     Using Ampere's Law, Magnetic field is given by B×2πr=2μor2a2$B\times 2\pi r={\displaystyle \frac{2{\mu }_o{r}^2}{a^2}}$
     B=μorπa2$B={\displaystyle \frac{{\mu }_{o}r}{\pi a^2}}$
For a≤r≤b$a\le r\le b$, 
     Current enclosed in the amperian loop is I2=2+3(r2−a2)b2−a2$I_2=2+{\displaystyle \frac{3(r^2-a^2)}{b^2-a^2}}$
     Using Ampere's Law, Magnetic field is given by B×2πr=μo(2+3(r2−a2)b2−a2)$B\times 2\pi r={\mu }_o(2+{\displaystyle \frac{3(r^2-a^2)}{b^2-a^2}})$
     B=μo2πr(2+3(r2−a2)b2−a2)$B={\displaystyle \frac{{\mu }_o}{2\pi r}}(2+{\displaystyle \frac{3(r^2-a^2)}{b^2-a^2}})$
For r≥b$r\ge b$,
     Current enclosed in the amperian loop is I3=5$I_3=5$
     B=5μo2πr$B={\displaystyle \frac{5{\mu }_o}{2\pi r}}$

Example

Magnetic field due to an infinitely long cylindrical cable with current density varying with radius

If the current density as a function of distance 'r$r$' from the axis of a radially symmetrical parallel stream of electrons is given as j(r)=xb(α+1)μ0rα−1${\displaystyle j(r)=\frac{xb(\alpha +1)}{{\mu }_0}{r}^{{\displaystyle \alpha -1}}}$ if the magnetic induction inside the stream varies as B=brα$B=b{r}^{\alpha }$, where b$b$ and α$\alpha$ are positive constants. Find 'x$x$'
Using Ampere's Circuital Law,∮pB.dl=μoI${\oint }_{p}B.dl={\mu }_{o}I$               (Equation 1)
Total current  (I$I$)  flowing through the cross-section of radius r$r$,  I=∫r0j(r′)×A$I={\int }_0^{r}j(r^{\prime })\times A$
 I=∫r0xb(α+1)(r′)α−1μo2πr′dr′$I={\int }_0^r{\displaystyle \frac{xb(\alpha +1)(r^{\prime }{)}^{\alpha -1}}{{\mu }_o}}2\pi r^{\prime }d{r}^{\prime }$I=xb(α+1)μo2π∫r0(r′)αdr′$I={\displaystyle \frac{xb(\alpha +1)}{{\mu }_o}}2\pi {\int }_0^r(r^{\prime }{)}^{\alpha }d{r}^{\prime }$I=xb(α+1)μo2πrα+1α+1$I={\displaystyle \frac{xb(\alpha +1)}{{\mu }_o}}2\pi {\displaystyle \frac{r^{\alpha +1}}{\alpha +1}}$
I=bx2πrα+1μo$I={\displaystyle \frac{bx2\pi r^{\alpha +1}}{{\mu }_o}}$
∮pB.dl=∮pbrαdl=brα∮pdl${\oint }_{p}B.dl={\oint }_{p}b{r}^{\alpha }dl=b{r}^{\alpha }{\oint }_{p}dl$
∮pB.dl=brα2πr=2πbrα+1${\oint }_{p}B.dl=b{r}^{\alpha }2\pi r=2\pi b{r}^{\alpha +1}$ From equation 1 , 2πbrα+1=μobx2πrα+1μo$2\pi b{r}^{\alpha +1}={\mu }_o{\displaystyle \frac{bx2\pi r^{\alpha +1}}{{\mu }_o}}$Hence, x=1$x=1$

Example

Magnetic field due to straight current carrying wires and loops

Two long straight conductors carry currents 4A$4A$ and 2A$2A$ into the plane of paper. A circular path is imagined to be enclosing these currents. The value of ∮B →.dl→$\oint \overrightarrow{B}.\overrightarrow{dl}$ is given by: (The value of magnetic permeability of vacuum is μ0${\mu }_0$)Using ampere's circuital law,
The line integral of magnetic field in a closed loop equals μ0i${\mu }_{0}i$ , where i$i$ is the current enclosed by the loop
ϕB¯⋅dl¯=μ0ienclosed$\varphi \overline{B}\cdot \overline{dl}={\mu }_0{i}_{enclosed}$
            =μ0(4+2)$={\mu }_0(4+2)$
            =6μ0$=6{\mu }_0$

Example

Magnetic field due to more than one current carrying wires

AB is the line on which the magnetic field of both the wires will cancel out. On CD, it will be added.
field due to X wire at P=μo2π2id$P={\displaystyle \frac{{\mu }_o}{2\pi }}{\displaystyle \frac{2i}{d}}$ outside of page
field due to Y at P=μo2π2id$P={\displaystyle \frac{{\mu }_o}{2\pi }}{\displaystyle \frac{2i}{d}}$ inside of page
find direction using right hand thumb rule
∴$\therefore$ at any point P on line AB,
Field=μo2i2πd−μo2i2πd$Field={\displaystyle \frac{{\mu }_{o}2i}{2\pi d}}-{\displaystyle \frac{{\mu }_{o}2i}{2\pi d}}$
=$=$zero

Formula

Magnetic field due to a long straight current carrying wire in a medium other than air

The magnetic field of an infinitely long straight wire in a medium other than air is given as 
B=μ0μrI2πr$B={\displaystyle \frac{{\mu }_0{\mu }_{r}I}{2\pi r}}$
where μr${\mu }_r$ is the relative permeability of the medium.

Example

Magnetic field due to a moving ring with uniform charge density

A thin disc (or dielectric) having radius r$r$ and charge q$q$ distributed uniformly over the disc is rotated n$n$ rotations per second about its axis. Find the magnetic field at the centre of the discSurface charge density σ=qπa2$\sigma ={\displaystyle \frac{q}{\pi a^2}}$
Charge on the hypothetical ring =qπa22πxdx$={\displaystyle \frac{q}{\pi a^2}}2\pi xdx$
Frequency of rotation=n,T=1n$=n,T={\displaystyle \frac{1}{n}}$
dI=qT=q1n=nq$dI=\frac{q}{T}={\displaystyle \frac{q}{{\displaystyle \frac{1}{n}}}}=nq$
Magnetic field due to the element
dB=μ0dI2x=μ02xdxqna2(2x)=μ0qndxa2$dB=\frac{{\mu }_{0}dI}{2x}=\frac{{\mu }_{0}2xdxqn}{a^2(2x)}=\frac{{\mu }_{0}qndx}{a^2}$
B=∫dB=μ0qna2∫a0dx=μ0qna2[x]a0=μ0qna$B=\int dB=\frac{{\mu }_{0}qn}{a^2}{\int }_0^{a}dx=\frac{{\mu }_{0}qn}{a^2}[x{]}_0^a=\frac{{\mu }_{0}qn}{a}$

Example

Magnetic compass

Magnetic compass works on the directive property of a magnet. It's needle is magnetic and allowed to rotate in a horizontal plane. One end of the needle always point north and is painted red. It is used for navigation purposes.

Definition

Directions Using a Magnet

When a bar magnet is hanged from a thread it comes to rest on north and south direction.The end of the magnet that points towards North is called its North seeking end or the North pole of the magnet. The other end that points towards the South is called South seeking end or the South pole of the magnet.
Hence the magnets are used in compass to give the directions and this is the property of magnet that when ever it is suspended freely it will align itself in north-south direction irrespective of its shape.

BookMarks

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