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

Field at the centre, $$B1$$ $$=$$μ$$04$$π ×   $$2$$$$π$$inr  $$=$$  μ$$02$$ .  nirField at a distance h from the centre,$$B2$$ $$=$$μ$$04$$π .   $$2$$$$π$$$$nir2r2$$ $$+$$ $$h23/2$$  $$=$$  μ$$02$$ .  $$nir2r3$$ $$1+h2r23/2=B1$$  $$1+h2r2$$−$$3/2$$ $$=$$  $$B1$$  $$1$$−$$32$$.$$h2r2$$                                                    $$(By$$ binomial $$theorem)Hence$$ $$B2$$ is less than $$B1$$ by a fraction $$=$$ $$32$$  $$h2r2$$

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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