Question

→ A circular coil of 20 turns and mean radius 6 cm carries a current of 2 A.
calculate the magnitude of magnetic field at a point 5 cm from the centre and
along the axis.

Answer 1

Answer:

A circular coil of 20 turns and mean radius 6 cm carries a current of 2 A, the magnitude of the magnetic field at a point 5 cm from the center and  along the axis is $=1.92*10^{-4}T$

Explanation:

According to Biot-Savart law, the magnetic field at a point P due to circular coil of radius ($r$) and turns ($n$) carrying current ($I$) at an axial distance ($x$) from the center is given by the formula,

$B=\frac{\mu \circ \ n I}{2}\frac{r^2}{\left(r^2+x^2\right)^{\frac{3}{2}}}$

From the question, it is given that

number of turns of the coil is $n=20$

the radius of the coil $r=6cm=0.06m$

current in the coil $I=2A$

the distance between the center of the coil and point P $x=5cm=0.05m$

substituting these values to get magnetic field,

$B=\frac{4\pi \times 10^{-7}\times 20\times 2}{2}\frac{0.06^2}{\left(0.06^2+0.05^2\right)^{\frac{3}{2}}}$

$=1.92*10^{-4}T$