Question

Applying biot-savart's law deduce the expression for the magnetic field at the centre of a semicircular loop of radius R carries current I

Answer 1

Applying biot-savart's, the expression for the magnetic field at the center of a semicircular loop of radius R carries a current I is μ₀NI/ 4R.

• The Magnetic field at the center due to a single current element of wire will be:  

       dB =  μ₀I dl sin 90° / 4πR²

             = μ₀I dl / 4πR²

• Therefore, the total magnetic field due to all current elements will be:

        B = ∫ dB

           = ∫ (μ₀I dl / 4πR²)

           = (μ₀I / 4πR²) ∫ dl

           = (μ₀I / 4πR²) x I

           = (μ₀I / 4πR²) x πR

           = μ₀I / 4R

• For N number of turns of the coil,

        B = μ₀NI / 4R

Answer 2

Expression for the magnetic field using Biot-Savart's law:

Applying Biot-Savart's law, the magnetic field in the current element dI is given as:

$\overrightarrow{\mathrm{dB}}=\frac{\mu_{0} I}{4 \pi} \cdot \frac{\overrightarrow{\mathrm{dl}} \times \vec{r}}{r^{3}}$

Where,

μ₀ = Permeability of free space

Consider a circular loop with 'N' number of turn and radius 'r'. The current flowing through the loop is 'I'.

Let's consider current element dl, the current element dl is perpendicular to position vector r.

dl ⊥ r

The magnetic field at the center of the loop is given as:

dB = μ₀I/4π (dl sin 90°)/r³

⇒ dB = μ₀I/4π (dl)/r³

The magnetic field due to current element point into center of the plane.

B = ∫ dB = ∫(μ₀Idl/4πr²)

B = μ₀I/4πr² ∫dl

B = μ₀I/4πr² × l

B = μ₀I/4πr² × 2πr

∴ B = μ₀I/2r

The magnetic field due to 'N' number of coils is given as:

∴ B = μ₀IN/2r