The magnetic field. A moving charge or an electric current sets up a magnetic field $\vec{B}$ in the space around it. Unlike the electric field, which is produced by charges at rest, the magnetic field is produced only by charges in motion. Its SI unit is the tesla (T), where $1\,\text{T} = 1\,\text{N A}^{-1}\text{m}^{-1}$; a smaller unit is the gauss, $1\,\text{G} = 10^{-4}\,\text{T}$.
Biot-Savart law. This law gives the small field $d\vec{B}$ produced by a current element $I\,d\vec{l}$ at a point P at distance $r$ from it. Its magnitude is $dB=\frac{\mu_0}{4\pi}\frac{I\,dl\sin\theta}{r^2}$, where $\theta$ is the angle between the element and the line joining it to P, and $\mu_0=4\pi\times10^{-7}\,\text{T m A}^{-1}$ is the permeability of free space. In vector form $d\vec{B}=\frac{\mu_0}{4\pi}\frac{I\,d\vec{l}\times\hat{r}}{r^2}$, so $d\vec{B}$ is perpendicular to the plane of $d\vec{l}$ and $\hat{r}$.
- The field is maximum when $\theta=90^\circ$ (point on the perpendicular) and zero when $\theta=0^\circ$ or $180^\circ$ (point on the axis of the element).
- It is the magnetic analogue of Coulomb's law, but the field falls off as $1/r^2$ from a current element while being directional.
Field due to a long straight wire. Integrating the Biot-Savart law for an infinitely long straight conductor gives $B=\frac{\mu_0 I}{2\pi a}$ at a perpendicular distance $a$. The lines are concentric circles around the wire; their direction is given by the right-hand thumb rule (thumb along current, curled fingers along $\vec{B}$).
Field on the axis of a circular loop. For a loop of radius $R$ carrying current $I$, the field on its axis at distance $x$ from the centre is $B=\frac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}$. At the centre $(x=0)$ this becomes $B=\frac{\mu_0 I}{2R}$; for $N$ turns, $B=\frac{\mu_0 N I}{2R}$.
Ampere's circuital law. The line integral of $\vec{B}$ around any closed loop equals $\mu_0$ times the total current enclosed: $\oint\vec{B}\cdot d\vec{l}=\mu_0 I_{enc}$. It is the magnetic counterpart of Gauss's law and is most useful for symmetric situations. Applying it to a straight wire instantly recovers $B=\frac{\mu_0 I}{2\pi a}$.
Solenoid and toroid. A solenoid is a long tightly wound coil. Using Ampere's law, the field deep inside a long solenoid is uniform: $B=\mu_0 n I$, where $n$ is the number of turns per unit length; outside it is nearly zero. A toroid is a solenoid bent into a closed ring; the field inside it is $B=\frac{\mu_0 N I}{2\pi r}$ and is confined entirely within the core, with no field outside.
A long straight wire carries a current of 5 A. Find the magnetic field at a perpendicular distance of 0.10 m from the wire.
Solution- Step 1: Use the long-straight-wire result $B=\frac{\mu_0 I}{2\pi a}$.
- Step 2: Substitute $\mu_0=4\pi\times10^{-7}$, $I=5$ A, $a=0.10$ m: $B=\frac{(4\pi\times10^{-7})(5)}{2\pi(0.10)}$.
- Step 3: Simplify: $B=\frac{2\times10^{-7}\times5}{0.10}=\frac{10^{-6}}{0.10}=1\times10^{-5}$ T.
Answer: $B=1\times10^{-5}\,\text{T}=10\,\mu\text{T}$, in concentric circles around the wire.
A circular coil of 50 turns and radius 0.05 m carries a current of 2 A. Find the magnetic field at its centre.
Solution- Step 1: Field at the centre of a coil: $B=\frac{\mu_0 N I}{2R}$.
- Step 2: Substitute $N=50$, $I=2$ A, $R=0.05$ m: $B=\frac{(4\pi\times10^{-7})(50)(2)}{2(0.05)}$.
- Step 3: Numerator $=4\pi\times10^{-7}\times100=4\pi\times10^{-5}$; denominator $=0.10$.
- Step 4: $B=\frac{4\pi\times10^{-5}}{0.10}=4\pi\times10^{-4}=1.26\times10^{-3}$ T.
Answer: $B\approx1.26\times10^{-3}\,\text{T}$ (about 12.6 G).
State the Biot-Savart law and write it in vector form. When is $dB$ maximum and when zero?
Solution- Step 1: Magnitude: $dB=\frac{\mu_0}{4\pi}\frac{I\,dl\sin\theta}{r^2}$, where $\theta$ is the angle between $d\vec{l}$ and $\hat{r}$.
- Step 2: Vector form: $d\vec{B}=\frac{\mu_0}{4\pi}\frac{I\,d\vec{l}\times\hat{r}}{r^2}$, so $d\vec{B}$ is perpendicular to both $d\vec{l}$ and $\hat{r}$.
- Step 3: $dB$ is maximum when $\theta=90^\circ$ ($\sin\theta=1$) and zero when $\theta=0^\circ$ or $180^\circ$ (point lies on the line of the element).
Answer: $dB=\frac{\mu_0}{4\pi}\frac{I\,dl\sin\theta}{r^2}$; maximum at $90^\circ$, zero along the element's own line.
A solenoid 0.5 m long has 1000 turns and carries a current of 3 A. Find the magnetic field inside it.
Solution- Step 1: Inside a long solenoid $B=\mu_0 n I$, where $n=N/L$.
- Step 2: $n=\frac{1000}{0.5}=2000$ turns/m.
- Step 3: $B=(4\pi\times10^{-7})(2000)(3)=4\pi\times10^{-7}\times6000$.
- Step 4: $B=24\pi\times10^{-4}=7.54\times10^{-3}$ T.
Answer: $B\approx7.54\times10^{-3}\,\text{T}$, uniform and directed along the axis.
Using Ampere's circuital law, derive the magnetic field at a distance $a$ from a long straight wire carrying current $I$.
Solution- Step 1: By symmetry $\vec{B}$ is the same magnitude on a circle of radius $a$ and is tangential to it.
- Step 2: Apply $\oint\vec{B}\cdot d\vec{l}=\mu_0 I_{enc}$ around this circle. Since $\vec{B}\parallel d\vec{l}$, the integral is $B(2\pi a)$.
- Step 3: The enclosed current is $I$, so $B(2\pi a)=\mu_0 I$.
- Step 4: Solve: $B=\frac{\mu_0 I}{2\pi a}$.
Answer: $B=\frac{\mu_0 I}{2\pi a}$, in agreement with the Biot-Savart result.
A toroid of mean radius 0.2 m has 4000 turns and carries 5 A. Find the field inside the core.
Solution- Step 1: Field inside a toroid: $B=\frac{\mu_0 N I}{2\pi r}$.
- Step 2: Substitute $N=4000$, $I=5$ A, $r=0.2$ m: $B=\frac{(4\pi\times10^{-7})(4000)(5)}{2\pi(0.2)}$.
- Step 3: Numerator $=4\pi\times10^{-7}\times20000=8\pi\times10^{-3}$; denominator $=0.4\pi$.
- Step 4: $B=\frac{8\pi\times10^{-3}}{0.4\pi}=2\times10^{-2}$ T.
Answer: $B=2\times10^{-2}\,\text{T}$, confined within the toroid core.