Question

Question 4.16: For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its centre is given by, (a) Show that this reduces to the familiar result for field at the centre of the coil. (b) Consider two parallel co-axial circular coils of equal radius R, and number of turns N, carrying equal currents in the same direction, and separated by a distance R. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to R, and is given by, , approximately. [Such an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.]

Class 12 - Physics - Moving Charges And Magnetism Moving Charges And Magnetism Page-170

Answer 1

on the aixs of the circular coil of radius R and N turns carrying a current I at a distance x from centre the magnetic field is given by
$B=\frac{\mu_0IR^2N}{2(x^2+R^2)^{3/2}}$

(a) at centre of the circular coil , x = 0
then, magnetic field,
$B=\frac{\mu_0IR^2N}{2R^3}=\frac{\mu_0NI}{2R}$
it is same result as is obtained by integration at the centre of coil.

(b) The side of the loop where current flows clockwise becomes South Pole and where the current appear anticlockwise becomes North Pole . direction of magnetic field is from South Pole to North Pole .
now, two parallel coils each of radius R with N turns . the magnetic field of both the coils add.

Net magnetic field , Bnet = B1 + B2
$B_{net}=\frac{\mu_0IR^2N}{2[(R/2)^2+R^2]^{3/2}}+\frac{\mu_0IR^2N}{2[(R/2)^2+R^2]^{3/2}}\\\\=\frac{2\mu_0IR^2N}{2[(5R^2/4)]^{3/2}}\\\\=\frac{\mu_0IR^2N}{2}\left[\begin{array}{c}\frac{8\times2}{5\sqrt{5}R^3}\end{array}\right]\\\\=\frac{8\mu_0NI}{5\sqrt{5}R}=0.72\frac{\mu_0NI}{R}$