Generalization Of Ampere’S Law And Its Applications - Ampere’S Law For More Than One Current Carrying Conductor

Generalization of Ampere’s law and its applications - Ampere’s law for more than one current carrying conductor

Ampere’s law can be extended to cases where there are multiple current carrying conductors
The net magnetic field at any point due to multiple currents is the vector sum of the individual magnetic fields at that point
The magnetic field at a point due to a particular current is given by the Ampere’s law equation

Statement of the generalized Ampere’s law

The line integral of the magnetic field around any closed path is equal to the sum of the products of the current and the element of the path
Mathematically, it can be written as: $$\oint \vec{B} \cdot d\vec{L} = \mu_0 \sum_{i} I_i$$ where:
$\oint \vec{B} \cdot d\vec{L}$ is the line integral of the magnetic field $\vec{B}$ around the closed path
$d\vec{L}$ is the element of the path
$\mu_0$ is the permeability of free space
$\sum_{i} I_i$ is the sum of the products of the current $I_i$ and the element of the path

Magnetic field due to two parallel current carrying conductors

Consider two parallel current carrying conductors carrying currents $I_1$ and $I_2$ respectively
The conductors are separated by a distance $d$
The magnitude of the magnetic field at a point between the conductors is given by Ampere’s law as: $$B = \frac{\mu_0}{2\pi} \cdot \frac{I_1 + I_2}{d}$$ where:
$B$ is the magnetic field at the point between the conductors
$\mu_0$ is the permeability of free space
$I_1$ and $I_2$ are the currents in the conductors
$d$ is the separation between the conductors

Magnetic field inside and outside a current carrying solenoid

Consider a long solenoid with $n$ turns per unit length and carrying a current $I$
The magnetic field inside the solenoid is: $$B = \mu_0 \cdot n \cdot I$$ where:
$B$ is the magnetic field inside the solenoid
$\mu_0$ is the permeability of free space
$n$ is the number of turns per unit length
$I$ is the current in the solenoid

Magnetic field outside a current carrying solenoid

The magnetic field outside the solenoid, at a distance $r$ from the solenoid, is nearly zero except in the vicinity of its ends
This is due to the cancellation of the magnetic fields produced by the current carrying loops
Near the ends of the solenoid, the magnetic field can be calculated using Ampere’s law

Magnetic field inside a toroid

A toroid is a hollow circular ring with a rectangular cross-section
It is typically wound with a coil carrying a current $I$
The magnetic field inside the toroid is: $$B = \mu_0 \cdot n \cdot I$$ where:
$B$ is the magnetic field inside the toroid
$\mu_0$ is the permeability of free space
$n$ is the number of turns per unit length
$I$ is the current in the toroid

Magnetic field outside a toroid

Outside the toroid, the magnetic field is practically zero
The magnetic field lines inside the toroid are predominantly confined within the toroid due to the shape and arrangement of the coil
This makes toroids useful in applications where we want to generate a strong and localized magnetic field

Application: Magnetic field of a current carrying wire

The magnetic field produced by a straight current carrying wire can be calculated using Ampere’s law
For a wire carrying a current $I$ at a distance $r$ from the wire, the magnetic field is given by: $$B = \frac{\mu_0 \cdot I}{2\pi r}$$ where:
$B$ is the magnetic field
$\mu_0$ is the permeability of free space
$I$ is the current in the wire
$r$ is the distance from the wire

Example: Magnetic field around a long straight wire

Consider a long straight wire carrying a current of 5 A
Calculate the magnetic field at a point which is at a distance of 10 cm from the wire Given:
Current $I = 5$ A
Distance $r = 10$ cm = 0.1 m Using the formula: $$B = \frac{\mu_0 \cdot I}{2\pi r}$$ Substituting the values: $$B = \frac{4\pi \times 10^{-7} \cdot 5}{2\pi \cdot 0.1}$$ Simplifying: $$B = 2 \times 10^{-6} \ T$$ Therefore, the magnetic field at the point is $2 \times 10^{-6}$ T.

Summary

Ampere’s law can be extended to cases involving multiple current carrying conductors
The line integral of the magnetic field around any closed path is equal to the sum of the products of the current and the element of the path
For two parallel current carrying conductors, the magnetic field at a point between them is given by: $B = \frac{\mu_0}{2\pi} \cdot \frac{I_1 + I_2}{d}$
Inside a solenoid or toroid, the magnetic field can be calculated using: $B = \mu_0 \cdot n \cdot I$
Outside a solenoid or toroid, the magnetic field is practically zero, except near the ends

Generalization of Ampere’s law and its applications - Ampere’s law for more than one current carrying conductor

Ampere’s law can be extended to cases where there are multiple current carrying conductors
The net magnetic field at any point due to multiple currents is the vector sum of the individual magnetic fields at that point
The magnetic field at a point due to a particular current is given by the Ampere’s law equation

Statement of the generalized Ampere’s law

The line integral of the magnetic field around any closed path is equal to the sum of the products of the current and the element of the path
Mathematically, it can be written as: $$\oint \vec{B} \cdot d\vec{L} = \mu_0 \sum_{i} I_i$$ where:
$\oint \vec{B} \cdot d\vec{L}$ is the line integral of the magnetic field $\vec{B}$ around the closed path
$d\vec{L}$ is the element of the path
$\mu_0$ is the permeability of free space
$\sum_{i} I_i$ is the sum of the products of the current $I_i$ and the element of the path

Magnetic field due to two parallel current carrying conductors

Consider two parallel current carrying conductors carrying currents $I_1$ and $I_2$ respectively
The conductors are separated by a distance $d$
The magnitude of the magnetic field at a point between the conductors is given by Ampere’s law as: $$B = \frac{\mu_0}{2\pi} \cdot \frac{I_1 + I_2}{d}$$ where:
$B$ is the magnetic field at the point between the conductors
$\mu_0$ is the permeability of free space
$I_1$ and $I_2$ are the currents in the conductors
$d$ is the separation between the conductors

Magnetic field inside and outside a current carrying solenoid

Consider a long solenoid with $n$ turns per unit length and carrying a current $I$
The magnetic field inside the solenoid is: $$B = \mu_0 \cdot n \cdot I$$ where:
$B$ is the magnetic field inside the solenoid
$\mu_0$ is the permeability of free space
$n$ is the number of turns per unit length
$I$ is the current in the solenoid

Magnetic field outside a current carrying solenoid

The magnetic field outside the solenoid, at a distance $r$ from the solenoid, is nearly zero except in the vicinity of its ends
This is due to the cancellation of the magnetic fields produced by the current carrying loops
Near the ends of the solenoid, the magnetic field can be calculated using Ampere’s law

Magnetic field inside a toroid

A toroid is a hollow circular ring with a rectangular cross-section
It is typically wound with a coil carrying a current $I$
The magnetic field inside the toroid is: $$B = \mu_0 \cdot n \cdot I$$ where:
$B$ is the magnetic field inside the toroid
$\mu_0$ is the permeability of free space
$n$ is the number of turns per unit length
$I$ is the current in the toroid

Magnetic field outside a toroid

Outside the toroid, the magnetic field is practically zero
The magnetic field lines inside the toroid are predominantly confined within the toroid due to the shape and arrangement of the coil
This makes toroids useful in applications where we want to generate a strong and localized magnetic field

Application: Magnetic field of a current carrying wire

The magnetic field produced by a straight current carrying wire can be calculated using Ampere’s law
For a wire carrying a current $I$ at a distance $r$ from the wire, the magnetic field is given by: $$B = \frac{\mu_0 \cdot I}{2\pi r}$$ where:
$B$ is the magnetic field
$\mu_0$ is the permeability of free space
$I$ is the current in the wire
$r$ is the distance from the wire

Example: Magnetic field around a long straight wire

Consider a long straight wire carrying a current of 5 A
Calculate the magnetic field at a point which is at a distance of 10 cm from the wire Given:
Current $I = 5$ A
Distance $r = 10$ cm = 0.1 m Using the formula: $$B = \frac{\mu_0 \cdot I}{2\pi r}$$ Substituting the values: $$B = \frac{4\pi \times 10^{-7} \cdot 5}{2\pi \cdot 0.1}$$ Simplifying: $$B = 2 \times 10^{-6} \ T$$ Therefore, the magnetic field at the point is $2 \times 10^{-6}$ T.

Summary

Ampere’s law can be extended to cases involving multiple current carrying conductors
The line integral of the magnetic field around any closed path is equal to the sum of the products of the current and the element of the path
For two parallel current carrying conductors, the magnetic field at a point between them is given by: $B = \frac{\mu_0}{2\pi} \cdot \frac{I_1 + I_2}{d}$
Inside a solenoid or toroid, the magnetic field can be calculated using: $B = \mu_0 \cdot n \cdot I$
Outside a solenoid or toroid, the magnetic field is practically zero, except near the ends

Calculation of magnetic field due to an arc of wire

To calculate the magnetic field due to an arc of wire, we can divide the arc into small elements
Each element produces a magnetic field at a point P
The magnetic field at P due to all the elements can be obtained by integrating the magnetic field produced by each element

Calculation of magnetic field due to an arc of wire (cont.)

The magnetic field at a point P due to a small element $d\vec{L}$ of wire can be given by the formula: $$d\vec{B} = \frac{\mu_0}{4\pi} \cdot \frac{I \cdot d\vec{L} \times \hat{r}}{r^2}$$ where:
$d\vec{B}$ is the magnetic field at point P due to the element $d\vec{L}$
$\mu_0$ is the permeability of free space
$I$ is the current in the wire
$r$ is the distance between the element and point P
$\hat{r}$ is the unit vector pointing from the element to point P

Calculation of magnetic field due to an arc of wire (cont.)

To find the magnetic field at point P due to the entire arc, we integrate the magnetic field produced by each element
The total magnetic field can be obtained by summing up the contributions from all the elements using the equation: $$\vec{B} = \oint \frac{\mu_0}{4\pi} \cdot \frac{I \cdot d\vec{L} \times \hat{r}}{r^2}$$ where the integration is performed over the entire arc

Example: Magnetic field due to a circular loop of wire

Consider a circular loop of wire with radius $R$ carrying a current $I$
We can calculate the magnetic field at the center of the loop using Ampere’s law
The magnetic field due to each small element $d\vec{L}$ of the loop is given by: $$d\vec{B} = \frac{\mu_0 \cdot I \cdot d\vec{L} \times \hat{r}}{4\pi R^2}$$ where $d\vec{L}$ is an infinitesimally small element of the loop, and $\hat{r}$ is the position vector pointing from the element to the center of the loop

Example: Magnetic field due to a circular loop of wire (cont.)

To find the total magnetic field at the center of the loop, we integrate the magnetic field produced by each element over the entire loop
The total magnetic field can be expressed as: $$\vec{B} = \oint \frac{\mu_0 \cdot I \cdot d\vec{L} \times \hat{r}}{4\pi R^2}$$ where the integration is performed over the entire loop

Example: Magnetic field due to a circular loop of wire (cont.)

For a circular loop of wire, the integration can be simplified by using the fact that the magnetic field produced by opposite elements in the loop cancel out
Only the magnetic field due to the current flowing in a semicircle of the loop contributes to the net magnetic field at the center
Therefore, we can consider only half of the loop in the integration

Example: Magnetic field due to a circular loop of wire (cont.)

For a circular loop of wire with radius $R$ carrying a current $I$, the magnetic field at the center of the loop can be calculated using: $$B = \frac{\mu_0 \cdot I}{2R}$$ where:
$B$ is the magnetic field at the center of the loop
$\mu_0$ is the permeability of free space
$I$ is the current in the loop
$R$ is the radius of the loop

Example: Magnetic field due to a circular loop of wire

Consider a circular loop of wire with radius 0.1 m carrying a current of 5 A
Calculate the magnetic field at the center of the loop Given:
Radius of the loop, $R = 0.1$ m
Current in the loop, $I = 5$ A Using the formula: $$B = \frac{\mu_0 \cdot I}{2R}$$ Substituting the values: $$B = \frac{4\pi \times 10^{-7} \cdot 5}{2 \cdot 0.1}$$ Simplifying: $$B = 1 \times 10^{-5} \ T$$ Therefore, the magnetic field at the center of the loop is $1 \times 10^{-5}$ T.

Summary

To calculate the magnetic field due to an arc of wire, we divide the arc into small elements and integrate the contributions from each element
The magnetic field at a point P due to a small element $d\vec{L}$ is given by: $d\vec{B} = \frac{\mu_0}{4\pi} \cdot \frac{I \cdot d\vec{L} \times \hat{r}}{r^2}$
The total magnetic field at P due to the entire arc is obtained by summing up the contributions from all the elements using: $\vec{B} = \oint \frac{\mu_0}{4\pi} \cdot \frac{I \cdot d\vec{L} \times \hat{r}}{r^2}$
For a circular loop of wire, the magnetic field at the center of the loop is given by: $B = \frac{\mu_0 \cdot I}{2R}$
The magnetic field due to opposite elements in the loop cancels out, and only the magnetic field due to the current flowing in