Physics HRK Notes: Ampere's Law & Biot-Savart Law - Complete Magnetic Field Guide

House of Physics October 01, 2025
Ampere's Law & Biot-Savart Law: Complete Guide to Magnetic Field Calculations
Complete Guide to Magnetic Field Calculations, Current Carrying Conductors & Applications - B.Sc. Physics Edition
Ampere's Law Biot-Savart Law Magnetic Field Calculations Solenoid Toroid Reading Time: 30 min

📋 Table of Contents

The Biot-Savart Law

🔬 Historical Background

Shortly after Oersted's discovery in 1819 that a compass needle is deflected by a current-carrying conductor, Jean-Baptiste Biot (1774–1862) and Félix Savart (1791–1841) performed quantitative experiments on the force exerted by an electric current on a nearby magnet.

📝 Experimental Observations of Biot and Savart

  • The vector \( d\mathbf{B} \) is perpendicular both to \( d\mathbf{s} \) (which points in the direction of the current) and to the unit vector \( \mathbf{\hat{r}} \) directed from \( d\mathbf{s} \) toward point \( P \).
  • The magnitude of \( d\mathbf{B} \) is inversely proportional to \( r^2 \), where \( r \) is the distance from \( d\mathbf{s} \) to \( P \).
  • The magnitude of \( d\mathbf{B} \) is proportional to the current \( I \) and to the magnitude \( |d\mathbf{s}| \) of the length element \( d\mathbf{s} \).
  • The magnitude of \( d\mathbf{B} \) is proportional to \( \sin\theta \), where \( \theta \) is the angle between the vectors \( d\mathbf{s} \) and \( \mathbf{\hat{r}} \).

📏 Mathematical Formulation

These observations are summarized in the mathematical expression known today as the Biot-Savart law:

\[ |d\mathbf{B}| \propto \frac{I \, ds \, \sin \theta}{r^2} \]
↓
\[ |d\mathbf{B}| = \frac{\mu_0 I}{4\pi} \frac{ds \, \sin \theta}{r^2} \]

Where \( \frac{\mu_0}{4\pi} \) is the constant of proportionality and \( \mu_0 = 4\pi \times 10^{-7} \, \text{Wb/A·m} \) is called the permeability of free space.

Vector Form of Biot-Savart Law

In vector form, the Biot-Savart law is expressed as:

\[ d\mathbf{B} = \frac{\mu_0 I}{4\pi} \frac{d\mathbf{s} \times \mathbf{\hat{r}}}{r^2} \]

Alternative Vector Form

Since \( \mathbf{r} = r\mathbf{\hat{r}} \), we can also write:

\[ d\mathbf{B} = \frac{\mu_0 I}{4\pi} \frac{d\mathbf{s} \times \mathbf{r}}{r^3} \]

Total Magnetic Field

The total magnetic induction \( \mathbf{B} \) at point \( P \) due to the entire wire is obtained by integration:

\[ \mathbf{B} = \int d\mathbf{B} = \frac{\mu_0 I}{4\pi} \int \frac{d\mathbf{s} \times \mathbf{r}}{r^3} \]

Applications of Biot-Savart Law

Magnetic Field due to Current in a Straight Conductor

🔌 Problem Setup

Consider a long straight conductor carrying current \( I \). We want to find the magnetic field strength at point \( P \) located at perpendicular distance \( a \) from the wire.

Element Selection

Consider a small length element \( ds \) at distance \( x \) from point \( O \) (taken as origin).

Biot-Savart Law Application

According to Biot-Savart law:

\[ dB = \frac{\mu_0}{4\pi} \frac{I \, ds \, \sin\theta}{r^2} \]

Geometric Relations

From geometry:

\[ r^2 = x^2 + a^2 \]
\[ ds = dx \]

Variable Transformation

Using trigonometric relations:

\[ \tan\theta = \frac{a}{-x} \Rightarrow x = -a \cot\theta \]
\[ dx = a \csc^2\theta \, d\theta \]

Substitution and Simplification

\[ dB = \frac{\mu_0}{4\pi} \frac{I (a \csc^2\theta \, d\theta) \sin\theta}{a^2 \cot^2\theta + a^2} \]
\[ = \frac{\mu_0}{4\pi} \frac{I (a \csc^2\theta \, d\theta) \sin\theta}{a^2 (1 + \cot^2\theta)} \]
\[ = \frac{\mu_0}{4\pi} \frac{I (a \csc^2\theta \, d\theta) \sin\theta}{a^2 \csc^2\theta} \]
\[ = \frac{\mu_0 I}{4\pi a} \sin\theta \, d\theta \]

Integration Limits

When \( x \to -\infty \): \( \theta \to 0 \)

When \( x \to \infty \): \( \theta \to \pi \)

Total Field Calculation

\[ B = \int dB = \frac{\mu_0 I}{4\pi a} \int_0^\pi \sin\theta \, d\theta \]
\[ = \frac{\mu_0 I}{4\pi a} [-\cos\theta]_0^\pi \]
\[ = \frac{\mu_0 I}{4\pi a} [-\cos\pi + \cos 0] \]
\[ = \frac{\mu_0 I}{4\pi a} [-(-1) + 1] \]
\[ = \frac{\mu_0 I}{4\pi a} (2) \]
\[ = \frac{\mu_0 I}{2\pi a} \]

📐 Final Result

The magnetic field due to a straight current-carrying conductor at distance \( a \) is:

\[ B = \frac{\mu_0 I}{2\pi a} \]

In vector form: \( \mathbf{B} = \frac{\mu_0 I}{2\pi a} \mathbf{\hat{n}} \), where \( \mathbf{\hat{n}} \) is the unit vector tangent to the circle at point P.

Magnetic Field due to a Circular Current Loop

🔄 Problem Setup

Consider a circular current loop of radius \( R \) carrying current \( I \). We want to find the magnetic field at point \( P \) located at distance \( x \) from the center of the loop.

Element Selection

Consider a small element of length \( ds \) such that the angle between \( \mathbf{r} \) and \( d\mathbf{s} \) is \( 90^\circ \).

Biot-Savart Law Application

According to Biot-Savart law:

\[ dB = \frac{\mu_0}{4\pi} \frac{I \, ds \, \sin 90^\circ}{r^2} = \frac{\mu_0 I}{4\pi} \frac{ds}{r^2} \]

Component Resolution

Resolve \( d\mathbf{B} \) into rectangular components \( dB_x \) and \( dB_y \). From symmetry, \( dB_y \) components cancel out, and only \( x \)-components contribute.

X-Component Integration

\[ B = \oint dB_x = \oint dB \cos\theta = \oint \frac{\mu_0 I}{4\pi} \frac{ds}{r^2} \cos\theta \]
\[ = \frac{\mu_0 I}{4\pi} \oint \frac{ds}{r^2} \cos\theta \]

Geometric Relations

\[ r^2 = x^2 + R^2 \]
\[ \cos\theta = \frac{R}{\sqrt{x^2 + R^2}} \]

Substitution

\[ B = \frac{\mu_0 I}{4\pi} \oint \frac{ds}{x^2 + R^2} \cdot \frac{R}{\sqrt{x^2 + R^2}} \]
\[ = \frac{\mu_0 I R}{4\pi (x^2 + R^2)^{3/2}} \oint ds \]

Integration Around Loop

\[ \oint ds = 2\pi R \]
\[ B = \frac{\mu_0 I R}{4\pi (x^2 + R^2)^{3/2}} (2\pi R) \]
\[ = \frac{\mu_0 I R^2}{2(x^2 + R^2)^{3/2}} \]

Case 1: Center of Loop

At the center of the loop, \( x = 0 \):

\[ B = \frac{\mu_0 I}{2R} \]

Case 2: Large Distance

At very large distance from the center (\( x \gg R \)):

\[ B = \frac{\mu_0 I R^2}{2x^3} \]

Case 3: N-Turn Coil

For a coil with \( N \) turns:

\[ B = \frac{\mu_0 N I A}{2\pi x^3} = \frac{\mu_0 \mu}{2\pi x^3} \]

where \( \mu = NIA \) is the magnetic dipole moment.

Force between Long Parallel Current Carrying Conductors

🔗 Problem Setup

Consider two long, straight, parallel wires separated by distance \( a \), carrying currents \( I_1 \) and \( I_2 \), respectively.

Magnetic Field from Wire 2

Wire 2 creates a magnetic field at the location of wire 1:

\[ B_2 = \frac{\mu_0 I_2}{2\pi a} \]

Force on Wire 1

The magnetic force on wire 1 due to wire 2:

\[ \mathbf{F_1} = I_1 \mathbf{L} \times \mathbf{B_2} \]
\[ F_1 = I_1 L B_2 \sin 90^\circ = I_1 L B_2 \]
\[ = I_1 L \left( \frac{\mu_0 I_2}{2\pi a} \right) = \frac{\mu_0 I_1 I_2}{2\pi a} L \]

Force on Wire 2

Similarly, the force on wire 2 due to wire 1:

\[ F_2 = \frac{\mu_0 I_1 I_2}{2\pi a} L \]

📐 Force Direction

The forces are equal in magnitude and opposite in direction. If the currents are in the same direction, the wires attract each other. If the currents are in opposite directions, the wires repel each other.

Ampere's Circuital Law

📜 Statement of Ampere's Law

"The line integral of magnetic induction \( \mathbf{B} \) around any closed path is equal to \( \mu_0 \) times the total current enclosed by the path."

Mathematical Form

\[ \oint \mathbf{B} \cdot d\mathbf{s} = \mu_0 I \]

where \( I \) is the total current enclosed by the path.

Multiple Currents

For multiple current-carrying conductors:

\[ \oint \mathbf{B} \cdot d\mathbf{s} = \mu_0 \sum I \]

where \( \sum I \) is the algebraic sum of all currents enclosed by the path.

Integral Form of Ampere's Law

📝 Current Density Form

If the current is distributed continuously, we can express the enclosed current in terms of current density \( \mathbf{J} \):

\[ \oint \mathbf{B} \cdot d\mathbf{s} = \mu_0 \int_S \mathbf{J} \cdot d\mathbf{A} \]

where \( S \) is any surface bounded by the closed path.

Differential Form of Ampere's Law

Applying Stokes' Theorem

Using Stokes' theorem:

\[ \oint_C \mathbf{B} \cdot d\mathbf{s} = \int_S (\nabla \times \mathbf{B}) \cdot d\mathbf{A} \]

Equating with Ampere's Law

\[ \int_S (\nabla \times \mathbf{B}) \cdot d\mathbf{A} = \mu_0 \int_S \mathbf{J} \cdot d\mathbf{A} \]

Differential Form

Since this holds for any surface \( S \), we have:

\[ \nabla \times \mathbf{B} = \mu_0 \mathbf{J} \]

📐 Comparison with Electrostatics

Ampere's law for magnetostatics is analogous to Gauss's law for electrostatics:

\[ \oint \mathbf{E} \cdot d\mathbf{A} = \frac{q}{\varepsilon_0} \quad \text{(Gauss's Law)} \]
\[ \oint \mathbf{B} \cdot d\mathbf{s} = \mu_0 I \quad \text{(Ampere's Law)} \]

Applications of Ampere's Law

Magnetic Field due to a Solenoid

🧲 Solenoid Definition

A solenoid is a long wire wound in the form of a helix with closely spaced turns. When current flows through it, it produces a nearly uniform magnetic field inside.

Amperian Loop Selection

Choose a rectangular Amperian loop with sides parallel and perpendicular to the solenoid axis.

Line Integral Calculation

\[ \oint \mathbf{B} \cdot d\mathbf{s} = \int_a^b \mathbf{B} \cdot d\mathbf{s} + \int_b^c \mathbf{B} \cdot d\mathbf{s} + \int_c^d \mathbf{B} \cdot d\mathbf{s} + \int_d^a \mathbf{B} \cdot d\mathbf{s} \]

Field Contributions

Outside the solenoid, \( B \approx 0 \). Inside, \( B \) is uniform and parallel to the axis.

\[ \oint \mathbf{B} \cdot d\mathbf{s} = B \cdot L + 0 + 0 + 0 = BL \]

Enclosed Current

If \( n \) is the number of turns per unit length, the enclosed current is:

\[ I_{enc} = nLI \]

Applying Ampere's Law

\[ \oint \mathbf{B} \cdot d\mathbf{s} = \mu_0 I_{enc} \]
\[ BL = \mu_0 nLI \]
\[ B = \mu_0 nI \]

📐 Final Result

The magnetic field inside a long solenoid is uniform and given by:

\[ B = \mu_0 nI \]

where \( n \) is the number of turns per unit length and \( I \) is the current.

Magnetic Field due to a Toroid

🍩 Toroid Definition

A toroid is a solenoid bent into the shape of a circle (doughnut shape). It has a uniform magnetic field inside the windings.

Amperian Loop Selection

Choose a circular Amperian loop inside the toroid, concentric with the toroid.

Symmetry Considerations

From symmetry, the magnetic field is constant in magnitude and tangential to the Amperian loop.

Line Integral Calculation

\[ \oint \mathbf{B} \cdot d\mathbf{s} = B \cdot 2\pi r \]

where \( r \) is the radius of the Amperian loop.

Enclosed Current

If \( N \) is the total number of turns and \( I \) is the current:

\[ I_{enc} = NI \]

Applying Ampere's Law

\[ \oint \mathbf{B} \cdot d\mathbf{s} = \mu_0 I_{enc} \]
\[ B \cdot 2\pi r = \mu_0 NI \]
\[ B = \frac{\mu_0 NI}{2\pi r} \]

📐 Field Variation

The magnetic field inside a toroid varies with the radial distance \( r \). It is maximum at the inner radius and minimum at the outer radius.

Frequently Asked Questions

❓ What is the difference between Biot-Savart Law and Ampere's Law?

Biot-Savart Law is used to calculate the magnetic field due to a small current element, while Ampere's Law relates the line integral of magnetic field around a closed loop to the current enclosed by that loop. Biot-Savart Law is more fundamental but Ampere's Law is often easier to apply in symmetric situations.

❓ When is Ampere's Law particularly useful?

Ampere's Law is particularly useful when the magnetic field has a high degree of symmetry, such as in the cases of long straight wires, solenoids, and toroids. In these cases, we can choose an Amperian loop where the magnetic field is either constant or zero along parts of the path.

❓ Why is the magnetic field outside a long solenoid approximately zero?

The magnetic field outside a long solenoid is approximately zero because the field lines are confined inside the solenoid due to the close spacing of turns. The field outside is the superposition of fields from adjacent turns that nearly cancel each other out.

❓ What is the significance of the permeability of free space μ₀?

The permeability of free space μ₀ is a fundamental physical constant that characterizes how a magnetic field interacts with matter. It appears in both Biot-Savart Law and Ampere's Law and relates magnetic fields to the currents that produce them. Its value is exactly 4π × 10⁻⁷ N/A².

❓ How does the magnetic field inside a toroid compare to that inside a solenoid?

Both solenoids and toroids produce confined magnetic fields, but the field inside a toroid is not uniform - it varies with the radial distance from the center. In contrast, the field inside a long solenoid is uniform. However, toroids have the advantage that there is no external magnetic field.

📚 Continue Your Physics Journey

Understanding Ampere's Law and Biot-Savart Law is crucial for mastering electromagnetism and its applications. These comprehensive notes provide a solid foundation for further studies in electromagnetic induction, AC circuits, and modern physics.

Explore More Topics of Electricity and Magnetism

© House of Physics | Chapter 35: Ampere's Law and Biot-Savart Law

These comprehensive notes are designed to help B.Sc. Physics students understand fundamental concepts of magnetic field calculations based on Halliday, Resnick and Krane

Author: Muhammad Ali Malik | Contact: +923016775811 | Email: aliphy2008@gmail.com

House of Physics

Posted by House of Physics

Post a Comment

0 Comments