Magnetism Concept Page - 5

Example

Magnetic field inside and outside a long straight current carrying wire

A long straight wire of a circular cross-section (radius a) carrying steady current I$I$. The current I$I$ is uniformly distributed across this cross-section. Calculate the magnetic field in the region r<a$r<a$ and r>a$r>a$ where a is the distance at which the magnetic field is calculated.

Solution (a) Consider the case r>a$r>a$. The outer Amperian loop,is a circle concentric with the cross-section. For this loop,
L=2πr$L=2\pi r$
Ie$I_e$ = Current enclosed by the loop = I$I$
The result is the familiar expression for a long straight wire
B(2πr)=μ0I$B(2\pi r)={\mu }_{0}I$B=μ0I2πr$B={\displaystyle \frac{{\mu }_{0}I}{2\pi r}}$
B∝1r   (r>a)$B\propto {\displaystyle \frac{1}{r}}(r>a)$. 

(b) Consider the case r<a$r<a$ i.e. the inner Amperian loop .For this loop, taking the radius of the circle to be r$r$, L=2πr$L=2\pi r$
Now the current enclosed Ie$I_e$ is not I$I$, but is less than this value.Since the current distribution is uniform, the current enclosed is,
Ie=I(πr2πa2)=Ir2a2$I_e=I({\displaystyle \frac{\pi r^2}{\pi a^2}})=I{\displaystyle \frac{r^2}{a^2}}$
Using Amperes law, B(2rπ)=μ0Ir2a2$B(2r\pi )={\mu }_{0}I{\displaystyle \frac{r^2}{a^2}}$
B=μ0Ir2πa2$B={\mu }_{0}I{\displaystyle \frac{r}{{2\pi a}^2}}$
B∝r   (r<a)$B\propto r(r<a)$

Example

Magnetic field due to infinite current carrying sheet

Consider an infinitely large sheet of thickness b$b$ lying in the xy-plane with a uniform current density J⃗ =J0i^$\overrightarrow{J}=J_0\hat{i}$ . Find the magnetic field everywhere.
Solution: We may think of the current sheet as a set of parallel wires carrying currents in the +x-direction.The magnetic field at a point P above the plane points in the positive y-direction. The z-component vanishes after adding up the contributions from all wires. Similarly, we may show that the magnetic field at a point below the plane points in the positive y-direction. We may now apply Amperes law to find the magnetic field due to the current sheet. 
Then
B=−μ0J0b2j^,z>b2$B=-{\displaystyle \frac{{\mu }_0{J}_{0}b}{2}}\hat{j},z>\frac{b}{2}$
=−μ0J0zj^,−b2<z<b2$=-{\mu }_0{J}_{0}z\hat{j},-{\displaystyle \frac{b}{2}}<z<\frac{b}{2}$
=μ0J0b2j^,z<−b2$={\displaystyle \frac{{\mu }_0{J}_{0}b}{2}}\hat{j},z<-{\displaystyle \frac{b}{2}}$

Definition

Calculate the magnitude of magnetic induction at a point on the axis at a distance from the centre of a current carrying coil

If there are n$n$ turns in the coil, then the magnitude of the magnetic induction at a point on the axis at a distance from the centre of current caring conductor
Bx=μ0nI2r2(x2+r2)(3/2)$B_x={\displaystyle \frac{{\mu }_{0}nI}{2}}{\displaystyle \frac{r^2}{{(x^2+r^2)}^{(}3/2)}}$
where μ0${\mu }_0$= the absolute permeability of free space.

Result

Magnetic field due to a circular loop

A circular loop behaves like a small electromagnet with pole faces as shown in the figure. This method to find the direction of magnetic poles is known as 'Clock Rule'.

Example

Magnetic field on the axis of a current carrying circular loop

The magnetic field at a distance x$x$ from the centre of the loop along the axial line is given as:
B=μ0IR22(x2+R2)32$B={\displaystyle \frac{{\mu }_{0}I{R}^2}{2(x^2+R^2{)}^{{\displaystyle \frac{3}{2}}}}}$
The magnetic induction at a point at a large distance d$d$ on the axial line of circular coil of small radius carrying current is 120$120$ μT$\mu T$. At a distance 2d$2d$ the magnetic induction would be:B=μ0⋅i⋅a22(a2+x2)32$B={\displaystyle \frac{{\mu }_0\cdot i\cdot a^2}{2(a^2+x^2{)}^{{\displaystyle \frac{3}{2}}}}}$
a<<<x$a<<<x$
so B=μ0ia22x3$B={\displaystyle \frac{{\mu }_{0}i{a}^2}{2x^3}}$
120=μia22d3$120={\displaystyle \frac{\mu i{a}^2}{2d^3}}$
so B1=μia22(2d)3$B^1={\displaystyle \frac{\mu i{a}^2}{2(2d{)}^3}}$
=μia22×8d3$={\displaystyle \frac{\mu i{a}^2}{2\times 8d^3}}$
=1208$={\displaystyle \frac{120}{8}}$
=15 μT$=15\mu T$

Example

Magnetic field at the centre of a circular current carrying loop

The magnetic field at the centre of a current carrying loop is:
B0=μ0I2R$B_0={\displaystyle \frac{{\mu }_{0}I}{2R}}$A circular coil of radius 25$25$ cm, carries a current of 50$50$ amperes. If it has 35$35$ turns, the flux density at the centre of the coil in Wb / m2$m^2$ is:B=nμ0i2r$B={\displaystyle \frac{n{\mu }_{0}i}{2r}}$
=35×4π×10−7×502×25×10−2$={\displaystyle \frac{35\times 4\pi \times {10}^{-7}\times 50}{2\times 25\times {10}^{-2}}}$
=1⋅4π×10−3T$=1\cdot 4\pi \times {10}^{-3}T$

Example

Magnetic Field due to various geometric configuration of circular loops

Example: Two identical circular loops each of radius r$r$ and carrying a current
i$i$ are arranged concentric with each other and in perpendicular planes as shown in the given figure. Find the magnetic field B⃗ $\overrightarrow{B}$ at the common center.

Solution:
Magnetic field due to a circular loop at the centre is: μ0i2r${\displaystyle \frac{{\mu }_{0}i}{2r}}$

|B|=B21+B22−−−−−−−√$\left|B\right|=\sqrt{B_1^2+B_2^2}$

|B|=(μ0i2r)2+(μ0i2r)2−−−−−−−−−−−−−√$\left|B\right|=\sqrt{({\displaystyle \frac{{\mu }_{0}i}{2r}}{)}^2+({\displaystyle \frac{{\mu }_{0}i}{2r}}{)}^2}$
= μ0ir2√ T${\displaystyle \frac{{\mu }_{0}i}{r\sqrt{2}}}T$

Example

Magnetic field due to concentric current carrying coils

Two concentric circular loops of radii r1$r_1$ and r2$r_2$ carry clockwise and anticlockwise currents i1$i_1$ and i2$i_2$ . If the the centre is a null point, then i1$i_1$/ i2$i_2$ ?B at centre μoi2r,${\displaystyle \frac{{\mu }_{o}i}{2r}},$  for zero magnetic field at centre,
∴$\therefore$ field produced by inner ring should be equal and opposite to that of ring 2.
μoI12r1=μoI22r2${\displaystyle \frac{{\mu }_o{I}_1}{2r_1}}={\displaystyle \frac{{\mu }_o{I}_2}{2r_2}}$
i1i2=r1r2

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