MP Board · Class 12 · Physics · Moving Charges and MagnetismState Biot-Savart Law. Derive an expression for the magnetic field at a point on the axis of a circular current-carrying loop. A circular coil of radius 10 cm, having 50 turns, carries a current of 2 A. Calculate the magnetic field at the center of the coil.

Biot-Savart Law and Magnetic Field Calculation

1. Biot-Savart Law Statement\nAccording to Biot-Savart Law, the magnetic field $dB$ at a point due to a current element $Idl$ carrying current $I$ at a distance $r$ from the element is directly proportional to:

• The current $I$
• The length of the current element $dl$
• The sine of the angle $\theta$ between the current element and the line joining the element to the point
• And is inversely proportional to the square of the distance $r$. \nMathematically: $$dB = \frac{\mu_0}{4\pi} \frac{I dl \sin\theta}{r^2}$$\nwhere $\mu_0$ is the permeability of free space.

2. Derivation of Magnetic Field on the Axis of a Circular Loop\nConsider a circular loop of radius $R$ carrying current $I$. Let $P$ be a point on its axis at a distance $x$ from the center $O$.

• Consider a small current element $dl$ at the top of the loop. The magnetic field $dB$ at point $P$ due to $dl$ is perpendicular to the line joining $dl$ and $P$.
• The distance $r = \sqrt{R^2 + x^2}$.
• By Biot-Savart law: $$dB = \frac{\mu_0}{4\pi} \frac{I dl \sin 90^\circ}{(R^2 + x^2)} = \frac{\mu_0}{4\pi} \frac{I dl}{R^2 + x^2}$$
• Resolving $dB$ into two components: $dB \sin\phi$ along the axis and $dB \cos\phi$ perpendicular to the axis.
• Due to symmetry, the perpendicular components ($dB \cos\phi$) cancel out, and only the axial components ($dB \sin\phi$) add up. \nTotal magnetic field $B$: $$B = \int dB \sin\phi = \int \frac{\mu_0}{4\pi} \frac{I dl}{R^2 + x^2} \cdot \frac{R}{\sqrt{R^2 + x^2}}$$ $$B = \frac{\mu_0 I R}{4\pi (R^2 + x^2)^{3/2}} \int dl = \frac{\mu_0 I R}{4\pi (R^2 + x^2)^{3/2}} (2\pi R)$$ $$B = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$$

3. Numerical Problem Solution

Given data:

• Radius of the coil ($R$) = $10 \text{ cm} = 0.1 \text{ m}$
• Number of turns ($N$) = $50$
• Current ($I$) = $2 \text{ A}$
• Permeability of free space ($\mu_0$) = $4\pi \times 10^{-7} \text{ T m A}^{-1}$

Formula:\nFor a coil with $N$ turns, the magnetic field at the center ($x = 0$) is given by: $$B = \frac{\mu_0 N I}{2 R}$$

Calculation:\nSubstitute the given values into the formula: $$B = \frac{(4\pi \times 10^{-7}) \times 50 \times 2}{2 \times 0.1}$$ $$B = \frac{4\pi \times 10^{-7} \times 100}{0.2}$$ $$B = \frac{4\pi \times 10^{-5}}{0.2}$$ $$B = 20\pi \times 10^{-5} \text{ T}$$ $$B = 2\pi \times 10^{-4} \text{ T} \approx 6.28 \times 10^{-4} \text{ T}$|

Answer:\nThe magnetic field at the center of the coil is $6.28 \times 10^{-4} \text{ Tesla}$.