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Magnetic field due to current element


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. 


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 cm

Now, consider only one side AB of the loop as shown in Fig. (b).

tanα=tanβ=(a/2)r=34 α=β=tan-134=37° Magnitude of magnetic induction at P, due to current in this side AB, is B0=μ0I4πr(sinα+sinβ)=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 2B0cosθ (along the axis).

Similarly, resultantof magnetic inductions produced by currents in remaining two opposite sides BC and AD will also be equal to 2B0cosθ (along the axis in same direction). Hence, resultant magnetic induction.

B=4B0cosθ B=4×3×10-6(a/2)r=9×10-6 T=9μT

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A small coil C with N= 200 turns is mounted on one end of a balance beam and introduced between the poles of an electromagnet as shown in figure. The area of the coil is S = 1cm2, the length of the right arm of the balance beam is I = 30 cm. When there is no current in the coil the balance is in equilibrium. On passing a current I = 22 mA through the coil, equilibrium is restored by putting an additional weight of mass m = 60 mg on the balance pan. Find the magnetic induction field (in terms of x 10-1 T) between the poles of the electromagnet, assuming it to be uniform.

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