# (a) Derive the expression for the magnetic field due to a current carrying coil of radius r at a distance x from the centre along the X-axis.(b) A straight wire carrying a current of 5 A is bent into a semicircular arc of radius 2 cm as shown in the figure. Find the magnitude and direction of the magnetic field at the centre of the arc.

(a )

Consider a circular loop of radius R carrying current i. We have to find the magnetic field at a point P on the axis of the loop at a distance x from the centre.  Now, consider a current element $id\vec l$ of the wire. The magnetic field at P due to this current element is :
$d\vec B=\frac{\mu_0}{4\pi}i\frac{d\vec l\times\vec r}{r^3}$
$d B=\frac{\mu_0}{4\pi}\frac{id l}{r^2}=\frac{\mu_0}{4\pi}\frac{id l}{\left ( R^2+X^2 \right )}$

The component along the x-axis is

The two field due to the opposite elements have a resultant along the axis of the loop. Dividing the loop into such a pair of diametrically opposite elements, the resultant field at P must be along the x axis  and vertical components get cancelled
So,
After integration,

(b)Due to straight wire portion magneticfield=0 and the magnetic field is only due to the semicircular arc.

$B=\frac{\mu _{0}I}{4r}$

$=\frac{4\pi \times 10^{-7}\times 5}{4\times 2\times10^{-2} }$

$=7.85\times 10^{-5}T$

The field is directed inwards perpendicular to the plane of the page.

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