Question

For a circular coil of radius R and N turns carrying current I, themagnitude of the magnetic field at a point on its axis at a distance xfrom its centre is given by,( )202 2 3/2 2IR NBx Rμ=+(a) Show that this reduces to the familiar result for field at thecentre 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 samedirection, and separated by a distance R. Show that the field onthe axis around the mid-point between the coils is uniform overa distance that is small as compared to R, and is given by,0 0.72NIBRμ= , approximately.[Such an arrangement to produce a nearly uniform magneticfield over a small region is known as Helmholtz coils.]

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Solution

The magnitude of magnetic field at the distance

Where,

By using Biot –Savart Law magnetic field

The perpendicular components of the field that is

From the

Then,

Now, on integrating the above equation over the circle of radius

The magnetic field due to a circular coil having

a)

The magnetic field at the centre of the circular coil is given as,

By substituting

b)

As the direction of the current flowing in the two coils is same so the two coils attract each other and the resultant field will be the sum of the field due to the coil

Where,

Therefore, we can write,

As

On solving further and taking

Thus magnetic field at the mid-point between the coils is uniform over a small distance

1

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