Question

A wire loop of the form shown in the figure carries a current I. Obtain the magnitude and direction of the magnetic field at P. 

Given : \(\cfrac{\mu _0 I}{4\pi R }\sqrt{2}\)

Answer 1

The wire loop is in the form of a circular arc AB of radius R and a straight conductor BCA. The arc AB subtends an angle of Φ = 270° = \(\cfrac{3\pi}{2}\)rad at the centre of the loop P. Since PA = PB = R and C is the midpoint of AB, AB =\(\sqrt{2}R\) and AC = CB =\(\cfrac{\sqrt{2}R}{2} = \cfrac{R}{\sqrt{2}}\). Therefore, a = PC = \(\cfrac{R}{\sqrt{2}}\).

The magnetic inductions at P due to the arc AB and the straight conductor BCA are respectively,

Therefore, the net magnetic induction at P is

This is the required expression.