NCERT · Class 12 · Physics · Moving Charges and MagnetismState Biot-Savart Law. Derive an expression for the magnetic field at the center of a circular current-carrying coil of radius $r$ having $N$ turns.

Biot-Savart Law

\nThe Biot-Savart Law is a fundamental law in electromagnetism that relates magnetic fields to the magnitudes, directions, length, and proximity of electric currents. According to this law, the magnitude of the magnetic field $dB$ at a point $P$ due to a current element $Idl$ 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 $P$
• And inversely proportional to the square of the distance $r$ from the current element to the point. \nMathematically, it is expressed as: $$dB = \frac{\mu_0}{4\pi} \frac{I dl \sin\theta}{r^2}$$\nwhere $\mu_0$ is the permeability of free space.

Derivation of Magnetic Field at the Center of a Circular Coil

\nLet us consider a circular coil of radius $r$ carrying a steady current $I$. Let the total number of turns in the coil be $N$.

1. Consider a small current element $dl$ on the circumference of the circular coil.
2. The angle $\theta$ between the current element $Idl$ and the position vector $r$ connecting the element to the center of the coil is always $90^\circ$ ($|dl \times r| = dlr \sin 90^\circ = dlr$).
3. Applying the Biot-Savart Law for this small element, the magnitude of the magnetic field $dB$ at the center is given by: $$dB = \frac{\mu_0}{4\pi} \frac{I dl \sin 90^\circ}{r^2} = \frac{\mu_0}{4\pi} \frac{I dl}{r^2}$|
4. The direction of this magnetic field, according to the right-hand thumb rule, is perpendicular to the plane of the coil and directed inwards.
5. To find the total magnetic field $B$ at the center due to the entire circular loop, we integrate $dB$ over the entire circumference: $$B = \int dB = \int \frac{\mu_0}{4\pi} \frac{I dl}{r^2}$$
6. Since $\frac{\mu_0}{4\pi}$, $I$, and $r^2$ are constants, they can be taken out of the integral: $$B = \frac{\mu_0 I}{4\pi r^2} \int dl$$
7. The integral of $dl$ over the entire circumference of the circle is equal to the total circumference, which is $2\pi r$: $$\int dl = 2\pi r$$
8. Substituting this value back into the equation: $$B = \frac{\mu_0 I}{4\pi r^2} (2\pi r) = \frac{\mu_0 I}{2r}$|
9. If the circular coil has $N$ turns, then the total magnetic field at the center is multiplied by $N$: $$B = \frac{\mu_0 N I}{2r}$$ \nThis is the required expression for the magnetic field at the center of a circular current-carrying coil.