Question

Obtain the formula for the magnetic field at the centre of a current carrying circular coil with the help of Biot-Savart law.

Hint:

For a coil of radius $R$ with $N$ turns, field at centre increases with $N$ and decreases with $R$.

Answer 1

Step 1: Biot–Savart law.
Magnetic field due to a small element $dl$ at point $P$: \[ dB = \frac{\mu_0}{4\pi} \frac{I \, dl \, \sin \theta}{r^2}. \]
Step 2: For circular loop.
At centre of coil: $r = R$, $\sin \theta = 1$, so \[ dB = \frac{\mu_0}{4\pi} \frac{I \, dl}{R^2}. \]
Step 3: Integrate over loop.
Total length of coil = $2 \pi R$: \[ B = \frac{\mu_0}{4\pi} \frac{I (2\pi R)}{R^2}. \] \[ B = \frac{\mu_0 I}{2R}. \]
Step 4: For N turns.
\[ B = \frac{\mu_0 N I}{2R}. \]
Step 5: Conclusion.
The magnetic field at the centre of circular coil is: \[ B = \frac{\mu_0 N I}{2R}. \]