Q&A - Ask Doubts and Get Answers
Q

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. Show that the field on the axis around the midpoint

b) 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,

 B = \frac{\mu _0 IR^2N}{2 ( x^2 + R^2 )^{3/2}} 

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,
B = 0.72 \frac{\mu _0 NI}{R}  approximately.

Answers (1)
Views
S Sayak

Let a point P be at a distance of l from the midpoint of the centres of the coils.

The distance of this point from the centre of one coil would be R/2+l and that from the other would be R/2-l.

The magnetic field at P due to one of the coils would be

.B_{1}= \frac{\mu _0 IR^2N}{2 ( (R/2+l)^2 + R^2 )^{3/2}}

The magnetic field at P due to the other coil would be

B_{2}= \frac{\mu _0 IR^2N}{2 ( (R/2-l)^2 + R^2 )^{3/2}}

Since the direction of current in both the coils is same the magnetic fields B1 and B2 due to them at point P would be in the same direction

Bnet =B1+B2

\\B_{net}= \frac{\mu _0 IR^2N}{2 ( (R/2-l)^2 + R^2 )^{3/2}}+\frac{\mu _0 IR^2N}{2 ( (R/2+l)^2 + R^2 )^{3/2}}\\ \\B_{net}= \frac{\mu _0 IR^2N}{2}\left [ ( (R/2-l)^2 + R^2 )^{-3/2} + ( (R/2+l)^2 + R^2 )^{-3/2}\right ]\\ \\B_{net}= \frac{\mu _0 IR^2N}{2}\left [ ( \frac{R^{2}}{4}-Rl+l^{2} + R^2 )^{-3/2} + ( \frac{R^{2}}{4}+Rl+l^{2} + R^2 )^{-3/2}\right ]\\ \\B_{net}= \frac{\mu _0 IR^2N}{2}\left [ ( \frac{5R^{2}}{4}-Rl+l^{2} )^{-3/2} + ( \frac{5R^{2}}{4}+Rl+l^{2} )^{-3/2}\right ]\\

\\B_{net}= \frac{\mu _0 IR^2N}{2}\times (\frac{5R^{2}}{4})^{-3/2}\left [ ( 1-\frac{4l}{5R}+\frac{4l^{2}}{5R^{2}} )^{-3/2} + ( 1+\frac{4l}{5R}+\frac{4l^{2}}{5R^{2}} )^{-3/2}\right ]\\

Since l<<R we can ignore term l2/R2

\\B_{net}= \frac{\mu _0 IN}{2R}\times (\frac{5}{4})^{-3/2}\left [ ( 1-\frac{4l}{5R} )^{-3/2} + ( 1+\frac{4l}{5R} )^{-3/2}\right ]\\

\\B_{net}= \frac{\mu _0 IN}{2R}\times (\frac{4}{5})^{3/2}\left [ 1+\frac{6l}{5R} + 1-\frac{6l}{5R} \right ]\\

\\B_{net}= \frac{\mu _0 IN}{2R}\times (\frac{4}{5})^{3/2}\times 2

\\B_{net}= 0.715\frac{\mu _0 IN}{R}\approx 0.72\frac{\mu _0 IN}{R}

Since the above value is independent of l for small values it is proved that about the midpoint the Magnetic field is uniform.

Exams
Articles
Questions