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Question
Using Biot-Savart law, show that magnetic flux density 'B' at the centre of a current carrying circular coil of radius R is given by: `B = ("μ"_0I)/(2R)` where the terms have their usual meaning.
Solution
Let a circular coil of radius R carry a current I and O be its centre.
By Biot-Savart's law, the magnitude of the magnetic field at O due a small element δ l of the loop is,
`δB = "μ"_0/(4π) (Iδl sinθ)/R^2`
Where θ is the angle between the length of the element (δ l) and the line joining the element to the point O. Here, θ = 90° because every element of. a circle is perpendicular to the radius.
∴ `δB = "μ"_0/(4π) (Iδl)/R^2`
Every component of the coil will have the same field direction. Because of the complete coil, the field `vec B` magnitude at O is:
`B = "μ"_0/(4pi)I/R^2 "∑"δl`
But `"∑ "δl = 2piR`
∴ `B = "μ"_0/(4pi) · I/R^2 xx 2piR`
or `B = ("μ"_0I)/(2R)`
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