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Question

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=μ0IR2N2(x2+R2)3/2

(a) Show that this reduces to the familiar result for field at the centre of the coil.

B=μ0IR2N2(x2+R2)3/2

(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=0.72μ0NIR

[Such an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.]

B=0.72μ0NIR

[Such an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.]

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Solution

(a)

At the centre of the coil, x=0.

Hence, magnetic field at centre is Bc=μoIR2N2R3

B=μoNI2R

In a small region of length 2d about the mid-point between the coils,

B=μ0IR2N2×⎡⎢⎣{(R2+d)2+R2}−3/2+{(R2−d)2+R2}−3/2⎤⎥⎦

=μ0IR2N2×(5R24)−3/2×[(1+4d5R)+(1−4d5R)−3/2]

=μ0IR2N2×(45)−3/2×[1−6d5R+1+6d5R]

where in the second and third steps above, terms containing d2/R2 and higher powers of d/R are neglected since dR<<1 The terms linear in d/R cancel giving a uniform field B in a small region:

B=μ0IR2N2×(45)−3/2μ0INR=0.72μ0INR

B=μ0IR2N2×⎡⎢⎣{(R2+d)2+R2}−3/2+{(R2−d)2+R2}−3/2⎤⎥⎦

=μ0IR2N2×(5R24)−3/2×[(1+4d5R)+(1−4d5R)−3/2]

=μ0IR2N2×(45)−3/2×[1−6d5R+1+6d5R]

where in the second and third steps above, terms containing d2/R2 and higher powers of d/R are neglected since dR<<1 The terms linear in d/R cancel giving a uniform field B in a small region:

B=μ0IR2N2×(45)−3/2μ0INR=0.72μ0INR

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