# Biot Savart Law Derivation

Created by: Team Physics - Examples.com, Last Updated: July 12, 2024

## Biot Savart Law Derivation

The Biot-Savart Law provides a way to calculate the magnetic field generated by a current-carrying conductor. It states that the infinitesimal magnetic field (ππ΅β) at a point in space due to a small segment of current (πΌ) is:

ππ΅β=πβ/4π πΌππβΓπβ/πΒ³

where:

• ππβ is the infinitesimal length vector of the current element.
• πβ is the position vector from the current element to the point where ππ΅β.
• π is the magnitude of πβ.
• π is the permeability of free space.

## Derivation

### Consider a Current Element:

Assume a small current element ππβ carrying a current πΌ.

### Apply the Concept of Magnetic Field:

The magnetic field due to this current element at a point π at a distance r is perpendicular to both the direction of the current and the line connecting the current element to the point π.

### Calculate the Magnetic Field:

The infinitesimal magnetic field is calculated by considering the contribution of the small current element using experimental observations and the cross product.

### Formulate the Biot-Savart Law:

By experimental measurements, it was found that ππ΅β is proportional to the current, ππβ, and sinβ‘π (where πΞΈ is the angle between ππβ and πβ), and inversely proportional to πΒ².

These findings form the basis of the Biot-Savart Law: ππ΅β=πβ/4ππΌβππβΓπβ/πΒ³β

## Example

Let’s consider an example of the Biot-Savart law to calculate the magnetic field at the center of a circular current-carrying loop.

### Magnetic Field at the Center of a Current-Carrying Loop

Given: A circular loop with radius π carrying a current πΌ.

Find: The magnetic field at the center of the loop.

### Solution:

Place the loop in the xy-plane with its center at the origin.

The current flows in a circular path in the counterclockwise direction.

For an infinitesimal current element ππβ on the loop, the position vector to the center of the loop is πβ, and π=π.

Since ππβ is tangential to the loop, πβ is perpendicular to ππβ.

The Biot-Savart Law for this current element becomes: ππ΅β=πβ/4ππΌβππβΓπβ/π3=πβ/4ππΌβππβ/πΒ²β

The cross product of ππβ and πβ simplifies because they are perpendicular, and the magnitude becomes ππββ1.

Since the magnetic field components due to each element ππβ are in the same direction (perpendicular to the loop plane), they add up constructively.

Integrating around the entire loop, the total magnetic field becomes: π΅=πβπΌ/4ππΒ²β2ππ=πβπΌ/2πβ

The factor 2π accounts for the total circumference of the loop.

The magnetic field at the center of a circular loop carrying current πΌ with radius π is π΅=πβπΌ/2R. This example shows how the Biot-Savart Law can be applied to find the magnetic field created by specific current distributions.

### Problem:

Find the magnetic field at the center of a square current-carrying loop with side length π and current πΌ.

### Solution:

The square loop lies in the xy-plane, centered at the origin.

Each side of the square contributes to the magnetic field at the center.

Applying Biot-Savart Law to One Side:

Consider one side of the loop parallel to the x-axis from βπ/2 to π/2.

The distance from each point on the side to the center is β(π/2)Β²+(π/2)Β²=π/β2β.

The magnetic field due to a segment ππ₯ is: ππ΅=πβπΌ/4πππ₯/(π/β2)Β²β

Summing Contributions from All Sides:

The total magnetic field is the vector sum of the contributions from all four sides.

The result is: π΅=2β2πβπΌ/ππβ.

## Practice Problems and Solutions

### Problem 1:

Calculate the magnetic field at a point on the axis of a circular loop of radius π, carrying a current πΌ, at a distance π₯ from the center of the loop.

### Solution:

Using Biot-Savart Law:

The magnetic field at a point on the axis is given by: ππ΅β=πβ/4ππΌβππβΓπβ/πΒ³β

ππβ is the small length element, and πβ is the distance from the element to the point on the axis.

Symmetry Considerations:

The tangential components cancel each other due to symmetry, and only the components along the axis contribute.

The total magnetic field along the axis (π΅π₯β) is given by: π΅π₯=πβπΌπΒ²/2(πΒ²+π₯Β²)^3/2

### Problem 2:

A straight conductor of length πΏ carries a current πΌ. Find the magnetic field at a point π perpendicular to the conductor, at a distance π from its midpoint.

### Solution:

Setup and Considerations:

Let the conductor lie along the x-axis from βπΏ/2 to πΏ/2.

Let the point π be along the y-axis at a distance π from the x-axis.

Applying the Biot-Savart Law:

The infinitesimal magnetic field due to an element ππ₯ at a distance

π=βπ₯Β²+πΒ²β is: ππ΅=πβπΌππ₯/4ππΒ²

The angle between ππβand πβ is 90β°, making the cross product ππβΓπβ=ππ₯.

Integrating to Find the Total Field:

Integrating from βπΏ/2 to πΏ/2, and considering only the perpendicular component: π΅=πβπΌπ/4πβ«βπΏ/2πΏ/2ππ₯(π₯Β²+πΒ²)^3/2β

The integral yields: π΅=πβπΌ/2ππ(πΏ/βπΏΒ²+4πΒ²)

### Problem 3:

Find the magnetic field at the center of a square loop of side length π, carrying current πΌ.

### Solution:

Analyzing the Problem:

The square loop can be divided into four equal sides.

By symmetry, each side contributes equally to the magnetic field at the center.

Applying the Biot-Savart Law:

Each side contributes a magnetic field perpendicular to the plane of the loop.

For each side, the magnetic field at the center is calculated using the Biot-Savart law:

ππ΅=πβπΌ/4πβ«βπ/2π/2ππ₯/(π/2)Β²

Combining Results:

After summing the contributions of all four sides: π΅=2β2πβπΌ/ππβ

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