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Moving Charges And Magnetism

Question
CBSEENPH12037584

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 = μ0IR2N2x2+R23/2
(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,
B = 0.72 μ0NIR, approximately.
[Such an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.]

Solution

(a) Given, a circular coil of radius r and N turns carrying current I.

Then,
Magnitude of magnetic field at a point on axis at a distance x from centre is given by,
            B = μ0IR2N2x2+R23/2   (axial line) 

At the centre of the coil x =0

Therefore,
 
                B = μ0IR2N2R3 

i.e.,           B = μ0IN2R 

which is same as the standard result. 

(b) In figure, O is a point which is mid-way between the two coils X and Y. 



Let, Bx be the magnetic field at Q due to coil X.
Then,
               Bx = μ0NIR22R2+d2+R23/2

If, By is the magnetic field at Q due to coil Y, then
              By = μ0NIR22R2-d2+R23/2 

The currents in both the coils X and Y are flowing in the same direction.  

So, the resultant field is given by