Ampere’s Law

Ampere's Law is a key rule in electromagnetism that links the magnetic field around a closed loop to the current passing through it. It says the magnetic field around the loop depends directly on the total current inside. This law is especially useful for calculating magnetic fields in simple, symmetrical setups like long wires, solenoids, and toroids. It’s an important part of Maxwell’s equations and helps explain how magnetic fields work in many situations.

1.0Definition of Ampere’s Law

• The line integral of the magnetic field B around any closed loop equals μ₀ multiplied by the net current enclosed within that loop.

$\oint \vec{B}\cdot d\vec{l}={\mu }_{0}i$

Ampere’s law provides a way to calculate the magnetic field from a current distribution using a closed planar loop (Amperian loop).

• Assign a direction to the loop by drawing an arrow. Using the right-hand rule, curl your fingers along the arrow; your thumb points to the positive side of the loop.
• The positive side can also be determined by looking along the loop’s axis: if the arrow is anticlockwise, the positive side faces you; if clockwise, it faces away.
• Consider a small element dl dl on the loop, and let $\vec{B}$ be the magnetic field there. The line integral (circulation) of $\vec{B}$ along the loop is: $B.dl\oint \vec{B}\cdot d\vec{l}$
• Currents crossing into the positive side are positive; those crossing out are negative.
• Ampere’s law states:

$\oint \vec{B}\cdot d\vec{l}={\mu }_0\times (\text{net current enclosed})$

• Only currents inside the loop affect the integral; external currents do not.
• The magnetic field $\vec{B}$ on the left side is the total field from all currents.
• Ampere’s law is equivalent to the Biot-Savart law but is simpler to apply in cases with high symmetry, making calculations easier.

2.0Application of Ampere’s Law

For Amperian Loop such that at each point of the loop, either

(1) B is tangential to the loop and is a non-zero constant B, or

(2) B is normal to the loop, or

(3) B vanishes

• Now, let L be the length (part) of the loop for which B is tangential.tangential. Let $I_e$ be the current enclosed by the loop. Then, by Ampere’s law

$BL={\mu }_0{I}_e$

3.0Magnetic Field of Infinite Straight Current-Carrying Conductor

$$\begin{gathered} \vec{B}\text{ will have circular lines.}\overrightarrow{dl}\text{is also taken tangent to the circle.} \\ \oint \vec{B}\cdot d\vec{l}=\oint Bdl \\ \therefore \theta =0^{\circ }\Rightarrow B\oint dl=B\times 2\pi R(\therefore B=\text{constant}) \\ \text{Now by Ampere’s law,} \\ B(2\pi R)={\mu }_{0}i \\ B=\frac{{\mu }_{0}i}{2\pi R} \end{gathered}$$

4.0Magnetic Field of Infinite Hollow Current-Carrying Cylinder

(I is uniformly distributed on the whole circumference)

1. For r ≥ R

Due to symmetry, the Amperian loop takes the shape of a circle.

$$\begin{gathered} \oint \vec{B}\cdot d\vec{l}=\oint Bdl \\ \therefore \theta =0^{\circ } \\ =B{\int }_0^{2\pi r}dl\therefore B=\text{Constant} \\ B=\frac{{\mu }_{0}i}{2\pi R} \\ \text{(2)}r<R \\ \oint \vec{B}\cdot d\vec{l}=\oint Bdl \\ =B(2\pi r)=0\Rightarrow B_{\text{in}}=0 \end{gathered}$$

5.0Magnetic Field of Infinite Solid Current-Carrying Cylinder

Assume the current is evenly distributed across the entire cross-sectional area.

Case (1) : r ≤ R

Consider an Amperian loop inside the cylinder. By symmetry, it should be a circle centered on the cylinder’s axis, with the loop’s axis aligned along the cylinder’s axis.

$$\begin{gathered} \oint \vec{B}\cdot d\vec{l}=\oint Bdl=B\oint dl=B\cdot 2\pi r={\mu }_0\frac{I}{\pi R^2}\pi r^2 \\ B=\frac{{\mu }_{0}Ir}{2\pi R^2}=\frac{{\mu }_{0}Jr}{2}\Rightarrow \vec{B}=\frac{{\mu }_0\vec{J}\times \vec{r}}{2} \end{gathered}$$

Case (2) : r ≥ R

$$\begin{gathered} \oint \vec{B}\cdot d\vec{l}=\oint Bdl=B\oint dl=B\cdot 2\pi r={\mu }_{0}I \\ \Rightarrow B=\frac{{\mu }_{0}I}{2\pi r} \\ \text{Also, }\vec{B}=\frac{{\mu }_{0}I}{2\pi r}\left(\vec{J}\times \vec{r}\right)=\frac{{\mu }_0\pi R^2}{2\pi r} \end{gathered}$$

6.0Non-Uniform Current Density

• Consider long, cylindrical conductor of radius R carrying a current I  with a non-uniform current density $J=\alpha r$ ,Where α is a constant.The problem can be solved by using the Ampere’s law.

$\oint \vec{B}\cdot d\vec{l}={\mu }_0\times I_{\text{enc}}$

$$\begin{gathered} \text{Where the enclosed current}{I}_{\text{enc}}\text{is given by} \\ {I}_{\text{enc}}=\int \vec{J}\cdot d\vec{A}=\int (\alpha r^{\prime })(2\pi r^{\prime }d{r}^{\prime }) \\ \text{(a) For r < R},theenclosedcurrentis \\ {I}_{\text{enc}}={\int }_0^{r}2\pi \alpha r^{\prime 2}d{r}^{\prime }=\frac{2\pi \alpha r^3}{3} \\ \text{Applying Ampere’s law, the magnetic field at}{P}_1\text{ is given by} \\ {B}_1(2\pi r)=\frac{2{\mu }_0\pi \alpha r^3}{3}\text{or}{B}_1=\frac{\alpha {\mu }_0}{3}{r}^2 \\ \text{This direction of the magnetic field}{\vec{B}}_1\text{is tangential to the Amperian loop which encloses the current. } \end{gathered}$$

(b) For r > R, the enclosed current is

$$\begin{gathered} I_{\text{enc}}={\int }_0^{R}2\pi \alpha r^{\prime 2}d{r}^{\prime }=\frac{2\pi \alpha R^3}{3} \\ \text{Which yields}{B}_2(2\pi r)=\frac{2{\mu }_0\pi \alpha R^3}{3} \\ \text{Thus, the magnetic field at a point}{P}_2\text{outside the conductor is,} \\ {B}_2=\frac{\alpha {\mu }_0{R}^3}{3r} \\ \text{The figure displays a plot of B versus r. } \end{gathered}$$

7.0Magnetic Field of Infinite Current-Carrying Sheet

• The figure shows an infinite current sheet with linear current density j (A/m). By symmetry, the magnetic field pattern above and below the sheet is uniform.
• Consider a square loop of side l

$$\begin{gathered} {\int }_a^b\vec{B}\cdot d\vec{l}+{\int }_b^c\vec{B}\cdot d\vec{l}+{\int }_c^d\vec{B}\cdot d\vec{l}+{\int }_d^a\vec{B}\cdot d\vec{l}={\mu }_0\cdot i(\text{By Ampere’s law}) \\ Since\vec{B}\perp d\vec{l}\text{along the path b}\to candd\to a,therefore{\int }_b^c\vec{B}\cdot d\vec{l}=0;{\int }_d^a\vec{B}\cdot d\vec{l}=0 \\ Also\vec{B}\parallel d\vec{l}\text{along the path a}\to \text{b and c}\to d,therefore{\int }_a^b\vec{B}\cdot d\vec{l}=0;{\int }_c^d\vec{B}\cdot d\vec{l}=2Bl \\ \text{The current confined by the loop is}i=jl \\ \text{Therefore, according to Ampere’s law}2Bl={\mu }_0\cdot jl\text{or}B=\frac{{\mu }_{0}j}{2} \end{gathered}$$

8.0Solenoid

• It is a coil of conducting wire wound tightly in a helical form around a cylindrical frame.
• The wire turns are electrically insulated from each other.
• It is used to produce a uniform and long-range magnetic field.
• The magnetic field along the axis is the result of the superposition of fields from many circular loops.
• These loops have their centers aligned along the solenoid’s axis.

9.0Magnetic field due to a Long Solenoid

A solenoid consists of a wire tightly wound into a helical coil.

• When the length of the solenoid is much greater than its radius r<<LIt is called an ideal solenoid.

For an ideal solenoid, the magnetic field inside is:

• Uniform,
• Parallel to the solenoid's axis,
• Strong in the central region.
• The magnetic field outside the solenoid is negligible (for an ideal case).
• The solenoid encompasses 'n' turns per unit length.
• The magnetic field can be calculated using Ampère’s circuital law.
• The field inside is produced due to the superposition of magnetic fields from each circular turn.      

$$\begin{gathered} B_{\text{inside}}=\frac{{\mu }_{0}ni}{2}\left[\sin 90^{\circ }+\sin 90^{\circ }\right]=\left({\mu }_{0}ni\right) \\ Also,{\int }_A^B\vec{B}\cdot d\vec{l}+{\int }_B^C\vec{B}\cdot d\vec{l}+{\int }_C^D\vec{B}\cdot d\vec{l}+{\int }_D^A\vec{B}\cdot d\vec{l}={\mu }_0\cdot i(\text{By Ampere’s law}) \\ {B}_{1}l+0+B_{2}l\cos 180^{\circ }=0^{\circ } \\ {B}_1-B_2=0 \\ {B}_1=B_2 \end{gathered}$$

Consider a rectangular loop ABCD. For this loop ${\oint }_{\text{ABCD}}\vec{B}\cdot d\vec{l}={\mu }_0{i}_{\text{enc}}$

$$\begin{gathered} {\oint }_{\text{ABCD}}\vec{B}\cdot d\vec{l}={\oint }_{\text{AB}}\vec{B}\cdot d\vec{l}+{\oint }_{\text{BC}}\vec{B}\cdot d\vec{l}+{\oint }_{\text{CD}}\vec{B}\cdot d\vec{l}+{\oint }_{\text{DA}}\vec{B}\cdot d\vec{l}=B\times a \\ \text{This is because} \\ {\oint }_{\text{AB}}\vec{B}\cdot d\vec{l}={\oint }_{\text{CD}}\vec{B}\cdot d\vec{l}=0,\vec{B}\perp d\vec{l} \\ {\oint }_{\text{DA}}\vec{B}\cdot d\vec{l}=0 \\ (\text{since }\vec{B}\text{ outside the solenoid is negligible}) \\ Now, \\ {i}_{\text{enc}}=(n\times a)\times i=B\times a={\mu }_0(n\times a\times i) \\ \Rightarrow B={\mu }_{0}ni \end{gathered}$$

For ideal solenoid:

1. MFI inside is uniform and axial

2. MFI outside solenoid is zero

10.0Toroid

• A toroid can be viewed as a solenoid bent into a circular ring, forming a closed loop. In this sense, it resembles an infinite cylindrical solenoid without ends.

Consider a toroid having n  turns per unit length. Magnetic field at a point P in the figure is given as

$B=\frac{{\mu }_{0}Ni}{2\pi r}={\mu }_{0}ni\text{where }n=\frac{N}{2\pi r}$

Illustration-1.B inside a cylindrical cavity inside a solid cylinder, at P ?

Solution:

$$\begin{gathered} B=\frac{{\mu }_0}{2}\vec{J}\times \overrightarrow{OP}+\frac{{\mu }_0}{2}\left(-\vec{J}\times \overrightarrow{O^{\prime }P}\right) \\ B=\frac{{\mu }_0}{2}\vec{J}\times \overrightarrow{OP}+\frac{{\mu }_0}{2}\left(\overrightarrow{O^{\prime }P}+\overrightarrow{P{O}^{\prime }}\right) \\ B=\frac{{\mu }_0}{2}\vec{J}\times \left(\overrightarrow{O{O}^{\prime }}\right) \end{gathered}$$