The field normal to the plane of a wire of n turns and radius r which carries a current i is measured on the axis of the coil at a small distance h from the centre of the coil. This is smaller than the field at the centre by the fraction

Question

Infinity Learn · Accepted Answer

The magnetic field on the axis of a current i carrying coil of turns n, radius r and at a distance i from the centre of the $$coilB=$$μ$$04$$π×$$2$$$$π$$$$ir2r2+h23/2$$…………………(1)The$$ field at the centre is given byBcentre $$=$$μ$$04$$π×$$2$$$$π$$ir…………………….$$(2)                                    $$($$∵ At centre $$h=0)BBcentre$$ $$=r3r2+h23/2=r3r31+h2r23/2=11+32h2r2$$ or $$B1+32h2r2=Bcentre$$∴ Bcentre −$$B/B=32h2r2$$

The field normal to the plane of a wire of n turns and radius r which carries a current i is measured on the axis of the coil at a small distance h from the centre of the coil. This is smaller than the field at the centre by the fraction

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