Question

Magnetic field on the axis of current carrying loop. Show its derivation.

Answer 1

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Answer 2

Magnetic field on the axis of current carring loop.
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see the figure ,
A current carring loop , axis OP
at point P , by elementary part of ring in both semicircle sides act magnetic field
dB .

In the figure it is clear that
dBcosα cancelled by other component dBcosα
hence,
∑B =2dBsinα
dB ={( μNidlsinθ)/4πr²}
so,
∑B =2 {( μNidlsinθ)/4πr²}sinα
= 2μNi.dl.sinθ/4πr².sinα
but for ring θ = 90°
∑B = 2μNi.dlsinα/4πr²

sinα = R/r but r =√(R² + x²)
so,
∑B = 2μNi.dlR/4π(R²+ x²)^3/2
= 2μNiR/4π(R²+ x²)^3/2∫dl
= μNiR²/2(R²+ x²)^3/2

Hence,
net magnetic field along axis , distance from centre of ring or loop is x and radius of loop is R , is μNiR²/2(R²+ x²)^3/2