A square loop of side a $$=$$ $$6$$ cm carries a current I $$=$$ $$1$$ A. Calculate magnetic induction B $$($$ in μT $$)$$ at point P, lying on the axis of loop and at a distance $$x=7$$ cm from the center of loop.

Question

A square loop of side a $$=$$ $$6$$ cm carries a current I $$=$$ $$1$$ A. Calculate magnetic induction B $$($$ in μT $$)$$ at point P, lying on the axis of loop and at a distance $$x=7$$ cm from the center of loop.

Infinity Learn · Accepted Answer

Axis of the loop means a line passing through centre of loop and normal to its plane. Since distance of the point P is x from centre of loop and side of square loop is a as shown in Fig. (a).Therefore, perpendicular distance of P from each side of the loop $$isr=x2+a22=4$$ cmNow, consider only one side AB of the loop as shown in Fig. (b).$$tan$$α$$=$$$$tan$$β$$=(a/2)r=34$$ α$$=$$β$$=$$$$tan$$$$-134=37$$° Magnitude of magnetic induction at P, due to current in this side AB, is $$B0=$$μ$$0I4$$$$π$$$$r($$$$sin$$α$$+$$$$sin$$β$$)=3$$×$$10-6$$ T Components of these magnetic inductions, parallel to plane of loop neutralise each other. Hence, resultant of these two magnetic inductions is $$2B0cos$$θ (along$$ the $$axis).Similarly, resultantof magnetic inductions produced by currents in remaining two opposite sides BC and AD will also be equal to $$2B0cos$$θ (along$$ the axis in same $$direction). Hence, resultant magnetic induction.$$B=4B0cos$$θ $$B=4$$×$$3$$×$$10-6(a/2)r=9$$×$$10-6$$ $$T=9$$μT

A square loop of side a = 6 cm carries a current I = 1 A. Calculate magnetic induction B ( in μT ) at point P, lying on the axis of loop and at a distance x=7 cm from the center of loop.

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