with the help of biot savarts law derive expression for magnetic field due to straight wire of length infinite carrying charge i

1. Break Down the Problem

To derive the expression for the magnetic field due to an infinite straight wire carrying a current $I$, we will:

1. Use Biot-Savart's Law.
2. Define appropriate variables.
3. Integrate to find the total magnetic field.

2. Relevant Concepts

Biot-Savart's Law states that the magnetic field $d\mathbf{B}$ at a point in space due to a differential length $d\mathbf{l}$ of a wire carrying a current $I$ is given by:

$d\mathbf{B} = \frac{\mu_0}{4\pi} \frac{I\, d\mathbf{l} \times \mathbf{r}}{r^3}$

Where:

• $\mu_0$ is the permeability of free space.
• $\mathbf{r}$ is the vector from the wire to the point where the magnetic field is being calculated.

3. Analysis and Detail

Let’s derive the magnetic field at a perpendicular distance $r$ from an infinite straight wire:

1. Setting the Coordinates:

   • Assume the wire lies along the z-axis, and we want to find the magnetic field at a point P, located at a distance $r$ from the wire along the x-axis.

2. Define $d\mathbf{l}$ and $\mathbf{r}$:

   • The differential length vector $d\mathbf{l}$ can be expressed as $d\mathbf{l} = dz\, \hat{k}$.
   • The position vector $\mathbf{r}$ from the wire to the point P is $\mathbf{r} = r\, \hat{i} - z\, \hat{k}$.

3. Calculating $r$ and $|\mathbf{r}|$:

   • The magnitude of $\mathbf{r}$ is given by: $|\mathbf{r}| = \sqrt{r^2 + z^2}$

4. Computing the Cross Product:

   • The cross product $d\mathbf{l} \times \mathbf{r}$: $d\mathbf{l} \times \mathbf{r} = dz\, \hat{k} \times (r\, \hat{i} - z\, \hat{k}) = dz\, (r\, \hat{j}) = r\, dz\, \hat{j}$

5. Substituting into Biot-Savart Law: $d\mathbf{B} = \frac{\mu_0}{4\pi} \frac{I\, r\, dz\, \hat{j}}{(r^2 + z^2)^{3/2}}$

6. Integrating Over the Wire:

   • The total magnetic field $\mathbf{B}$ is found by integrating from $-\infty$ to $+\infty$: $\mathbf{B} = \int_{-\infty}^{\infty} d\mathbf{B} = \frac{\mu_0 I r}{4\pi} \int_{-\infty}^{\infty} \frac{dz}{(r^2 + z^2)^{3/2}}$

7. Evaluate the Integral:

   • The integral can be solved, resulting in: $\int_{-\infty}^{\infty} \frac{dz}{(r^2 + z^2)^{3/2}} = \frac{2}{r^2}$

8. Final Substitution:

   • Therefore, substituting the integral back into the magnetic field expression: $\mathbf{B} = \frac{\mu_0 I r}{4\pi} \cdot \frac{2}{r^2} \hat{j} = \frac{\mu_0 I}{2\pi r} \hat{j}$

4. Verify and Summarize

Upon deriving, we find that the magnetic field $\mathbf{B}$ produced by an infinite straight wire carrying a current $I$ at a distance $r$ is:

$\mathbf{B} = \frac{\mu_0 I}{2\pi r} \hat{j}$

This formula indicates that the magnetic field circles around the wire and its magnitude decreases with increasing distance $r$.

Final Answer

The expression for the magnetic field due to an infinite straight wire carrying a current $I$ is:

$\mathbf{B} = \frac{\mu_0 I}{2\pi r} \hat{j}$