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प्रश्न
Derive an expression for axial magnetic field produced by current in a circular loop.
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उत्तर
- Consider loop of radius R carrying current I placed in x - y plane with its centre at origin O as shown in the figure below.
The magnetic field on the axis of a circular current loop of radius R - Let point P can be on z-axis at distance `vec"r"` from line element `vec"d"l` of the loop.
- Using Biot-Savart law, the magnitude of the magnetic field dB is given by dB = `mu_0/(4pi) "I" (|"d"vecl xx vec"r"|)/"r"^3`
- Any element `vec"d"l` will always be perpendicular to the vector `vec"r"` from the element to the point P. The element `vec"d"l` is in the x-y plane, while the vector `vec"r"` is in the y-z plane. Hence
`vec"d"l` × `vec"r"` = dlr
∴ dB = `mu_0/(4pi) "I" "dl"/"r"^2`
but r2 = R2 + z2
∴ dB = `mu_0/(4pi) "I" "dl"/(("z"^2 + "R"^2))` - Now, the direction of `"d"vec"B"` is perpendicular to the plane formed by `"d"vec"l"` and `vec"r"`. Its z component is dBz and the component perpendicular to the z-axis is dB⊥. The components dB⊥ when summed over, yield zero as they cancel out due to symmetry. Hence, only the z component remains.
- The net contribution along the z-axis is obtained by integrating dBz = dB cos θ over the entire loop. From figure,
cosθ = `"R"/"r" = "R"/sqrt("z"^2 + "R"^2)`
∴ Bz = `int "dB"_"z" = mu_0/(4pi) "I" int "dl"/(("z"^2 + "R"^2))cos theta`
= `mu_0/(4pi) "I" int "Rdl"/(("z"^2 + "R"^2)^{3/2})`
= `mu_0/(4pi) xx "lR"/(("z"^2 + "R"^2)^{3/2}) xx 2pi"R"`
Bz = `mu_0/2 xx "IR"^2/(("z"^2 + "R"^2)^{3/2})`
This is the magnitude of the magnetic field due to current I in the loop of radius R, on a point at P on the z-axis of the loop.
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