MP Board · Class 12 · Physics · Moving Charges and MagnetismState Biot-Savart law. Using this law, derive an expression for the magnetic field at a point situated at a distance $x$ from the center along the axis of a circular current-carrying coil of radius $R$ and having $N$ turns.

Biot-Savart Law

\nThe Biot-Savart law is a mathematical expression that describes the magnetic field generated by a constant electric current. It states that the magnetic field contribution ($d\vec{B}$) at a point $P$ due to a current element ($I d\vec{l}$) is directly proportional to the current ($I$), the magnitude of the length element ($dl$), the sine of the angle ($\theta$) between the length element vector and the displacement vector ($\vec{r}$) connecting the current element to the point, and inversely proportional to the square of the distance ($r$) from the current element to the point. \nMathematically, it is expressed as: $$d\vec{B} = \frac{\mu_0}{4\pi} \frac{I (d\vec{l} \times \vec{r})}{r^3}$|\nIn magnitude form: $$dB = \frac{\mu_0}{4\pi} \frac{I dl \sin\theta}{r^2}$$\nwhere $\mu_0$ is the permeability of free space.

Derivation of Magnetic Field on the Axis of a Circular Coil

1. Consideration: Consider a circular coil of radius $R$ carrying a steady current $I$ with $N$ turns. Let the axis of the coil be the X-axis. We need to find the magnetic field at point $P$ situated at a distance $x$ from the center $O$ along the axis.

2. Current Element: Consider a small current element $Idl$ at the top of the coil. The distance $r$ from this element to point $P$ is given by: $$r = \sqrt{R^2 + x^2}$$

3. Applying Biot-Savart Law: The angle between the current element $Id\vec{l}$ and position vector $\vec{r}$ is $90^\circ$. Therefore, the magnitude of the magnetic field $dB$ due to this element at point $P$ is: $$dB = \frac{\mu_0}{4\pi} \frac{I dl \sin(90^\circ)}{r^2} = \frac{\mu_0}{4\pi} \frac{I dl}{R^2 + x^2}$$

4. Resolving Components: The direction of $dB$ is perpendicular to the plane containing $Id\vec{l}$ and $\vec{r}$. Let $\alpha$ be the angle between the axis and the vector $dB$. Resolving $dB$ into two rectangular components:

   • Component along the axis: $dB \cos\alpha$
   • Component perpendicular to the axis: $dB \sin\alpha$

5. Symmetry Argument: For every current element at the top, there is a diametrically opposite element at the bottom whose perpendicular components ($dB \sin\alpha$) are equal in magnitude and opposite in direction, hence they cancel each other out. Only the axial components ($dB \cos\alpha$) contribute to the total magnetic field.

6. Integration: The total magnetic field $B$ is the sum of all axial components around the circular loop: $$B = \int dB \cos\alpha$$ From the geometry of the triangle, $\cos\alpha = \frac{R}{r} = \frac{R}{\sqrt{R^2 + x^2}}$ Substituting the values of $dB$ and $\cos\alpha$: $$B = \int \left( \frac{\mu_0}{4\pi} \frac{I dl}{R^2 + x^2} \right) \left( \frac{R}{\sqrt{R^2 + x^2}} \right)$$ $$B = \frac{\mu_0 I R}{4\pi (R^2 + x^2)^{3/2}} \int dl$$

7. Final Expression: Since the total length of the circular coil of $N$ turns is $\int dl = 2\pi R N$: $$B = \frac{\mu_0 I R}{4\pi (R^2 + x^2)^{3/2}} (2\pi R N)$$ $$B = \frac{\mu_0 I R^2 N}{2 (R^2 + x^2)^{3/2}}$$