Ncert Exercises & Solutions

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,

$B = \frac{μ_{0} {IR}^{2} N}{2 {(x^{2} + R^{2})}^{\frac{3}{2}}}$

(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, $B = 0.72 - \frac{μ_{0} BNI}{R}$ , approximately.

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

Radius of circular coi=R

Number of turns on the coil=N

Current in the coil=I

Magnetic field at a point on its axis at distance x is given by the relation,

B = \frac{μ_{0} {IR}^{2} N}{2 {(x^{2} + R^{2})}^{\frac{3}{2}}}

Where,

μ_{0} = Permeability of free space = 4 π \times 10^{- 7} T m A^{- 1}

(a) If the magnetic field at the centre of the coil is considered, then x=0.

∴ B = \frac{μ_{0} {IR}^{2} N}{2 R^{3}} = \frac{μ_{0} IN}{2 R}

This is the familar result for magnetic field at the centre of the coil.

(b) Radii of two parallel co-axial circular coils=R

Number of turns on each coil=N

Current in both coils=I

Distance between both the coils=R

Let us consider point Q at distance d from the centre.

Then, one coil is at a distance of

\frac{R}{2} + d

from point Q.

Magnetic field at point Q is given as:

∴ B_{1} = \frac{μ_{0} {NIR}^{2}}{2 {[{(\frac{R}{2} + d)}^{2} + R^{2}]}^{\frac{3}{2}}}

Also, the other coil is at a distance of

\frac{R}{2} - d

from point Q.

Magnetic field due to this coil is given as:

∴ B_{2} = \frac{μ_{0} {NIR}^{2}}{2 {[{(\frac{R}{2} - d)}^{2} + R^{2}]}^{\frac{3}{2}}}

Total magnetic field,

B = B_{1} + B_{2}

= \frac{μ_{0} {IR}^{2}}{2} [{{(\frac{R}{2} - d)}^{2} + R^{2}}^{\frac{3}{2}} + {{(\frac{R}{2} + d)}^{2} + R^{2}}^{\frac{3}{2}}]

= \frac{μ_{0} {IR}^{2}}{2} [{(\frac{5 R^{2}}{4} + d^{2} - Rd)}^{\frac{3}{2}} + {(\frac{5 R^{2}}{4} + d^{2} + Rd)}^{\frac{3}{2}}]

= \frac{μ_{0} {IR}^{2}}{2} \times {(\frac{5 R^{2}}{4})}^{\frac{3}{2}} [{(1 + \frac{4}{5} \frac{d^{2}}{R^{2}} - \frac{4}{5} \frac{d}{R})}^{\frac{3}{2}} + {(1 + \frac{4}{5} \frac{d^{2}}{R^{2}} + \frac{4}{5} \frac{d}{R})}^{\frac{3}{2}}]

For d < < R, neglecting the factor \frac{d^{2}}{R^{2}}, we get :

\approx \frac{μ_{0} {IR}^{2}}{2} \times {(\frac{5 R^{2}}{4})}^{\frac{3}{2}} \times [{(1 - \frac{4 d}{5 R})}^{\frac{3}{2}} + {(1 + \frac{4 d}{5 R})}^{\frac{3}{2}}]

\approx \frac{μ_{0} {IR}^{2}}{2} \times {(\frac{4}{5})}^{\frac{3}{2}} [1 - \frac{6 d}{5 R} + 1 + \frac{6 d}{5 R}]

B = {(\frac{4}{5})}^{\frac{3}{2}} \frac{μ_{0} IN}{R} = 0.72 (\frac{μ_{0} IN}{R})

Hence, it is proved that the field on the axis around the mid-point between the coils is uniform.

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