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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 = `(μ_0"IR"^2"N")/(2("x"^2 + "R"^2)^(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 (μ_0"NI")/"R"` approximately.
[Such an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.]
Solution
Radius of circular coil = 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 = `(μ_0"IR"^2"N")/(2("x"^2 + "R"^2)^(3/2))`
Where,
μ0 = Permeability of free space
(a) If the magnetic field at the centre of the coil is considered, then x = 0.
∴ B = `(μ_0"IR"^2"N")/(2"R"^3) = (μ_0"IN")/(2"R")`
This is the familiar 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 `"R"/2 + "d"` from point Q.
∴ Magnetic field at point Q is given as:
B1 = `(μ_0"NIR"^2)/(2[("R"/2 + "d")^2 + "R"^2]^(3/2))`
Also, the other coil is at a distance of `"R"/2 - "d"` from point Q.
∴ Magnetic field due to this coil is given as:
B2 = `(μ_0"NIR"^2)/(2[("R"/2 - "d")^2 + "R"^2]^(3/2))`
Total magnetic field,
B = B1 + B2
= `(μ_0"IR"^2)/2 [{("R"/2 - "d")^2 + "R"^2}^(3/2) + {("R"/2 + "d")^2 + "R"^2}^(3/2)]`
= `(μ_0"IR"^2)/2 [((5"R"^2)/4 + "d"^2 - "Rd")^(3/2) + ((5"R"^2)/4 + "d"^2 + "Rd")^(3/2)]`
= `(μ_0"IR"^2)/2 xx ((5"R"^2)/4)^(3/2) [(1 + 4/5 "d"^2/"R"^2 - 4/5 "d"/"R")^{3/2) + (1 + 4/5 "d"^2/"R"^2 + 4/5 "d"/"R")^(3/2)]`
For d << R, neglacting the factor `"d"^2/"R"^2`, we get:
`≈ (μ_0"IR"^2)/2 xx ((5"R"^2)/4)^(3/2) xx [(1 - (4"d")/(5"R"))^(3/2) + (1 + (4"d")/(5"R"))^(3/2)]`
`≈ (μ_0"IR"^2"N")/(2"R"^3) xx (4/5)^(3/2) [1 - (6"d")/(5"R") + 1 + (6"d")/(5"R")]`
B = `(4/5)^(3/2) (μ_0"IN")/"R" = 0.72 ((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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