(a) Using Biot-Savart's law, derive an expression for the magnetic field at the centre of a circular coil of radius R, number of turns N, carrying current.
(b) Two small identical circular coils marked 1,2 carry equal currents and are placed with their geometric axes perpendicular to each other as shown in the figure. Derive an expression for the resultant magnetic field at O.

Explanation

(a) According to Biot-Savart law, the magnetic field due to a current element `vec(dt)` at the observation point whose position vector `vec(r)` is given by
`vec(d B)=(mu_(0)I)/(4pi).(vec(dl)xx vec(r))/(r^(3))`
The direction of dl is along the tangent, so dl `_|_` r. From Biot-Savart law, magnetic field at the Centre O due to this current element is
`dB=(mu_(0)I)/(4pi)(dl sin 90^(@))/(r^(2))=(mu_(0)I)/(4pi) (dl)/(r^(2))`
The magnetic field due to all such current elements will point into the plane of paper at the centre O. Hence the total magnetic field at the centre O is
`B=int dB= int(mu_(0)I dl)/(4pi r^(2))=(mu_(0)I)/(4pi r^(2)) int dl = (mu_(0)I)/(4pi r^(2)).l`
For a coil of N turns, B `=(mu_(0)NI)/(2r)`.
(b) Magnetic field at O due to loop 1.
`B_(1)=(mu_(0)iR^(2))/(2(x^(2)+R^(2))^(3//2))`, acting towards left
Magnetic field at O due to loop 2.
`B_(2)=(mu_(0)iR^(2))/(2(x^(2)+R^(2))^(3//2))`, acting vertically upwards
Here, R is the radius of each loop.
Resultant field at O will be
`B=sqrt(B_(1)^(2)+B_(2)^(2))=sqrt(2)B_(1) (.:' B_(1)=B_(2))=(mu_(0))/(sqrt(2))(iR^(2))/((x^(2)+R^(2))^(3//2))`
This field acts at an angle of `45^(@)` with the axis of loop 1.