Question

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

Moderate

Solution

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 is$r=\sqrt{{x}^{2}+{\left(\frac{a}{2}\right)}^{2}}=4\mathrm{cm}$

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

$\mathrm{tan}\alpha =\mathrm{tan}\beta =\frac{(a/2)}{r}=\frac{3}{4}\phantom{\rule{0ex}{0ex}}\alpha =\beta ={\mathrm{tan}}^{-1}\left(\frac{3}{4}\right)=37\xb0\phantom{\rule{0ex}{0ex}}MagnitudeofmagneticinductionatP,duetocurrentinthissideAB,is\phantom{\rule{0ex}{0ex}}{B}_{0}=\frac{{\mu}_{0}I}{4\pi r}(\mathrm{sin}\alpha +\mathrm{sin}\beta )=3\times {10}^{-6}\mathrm{T}\phantom{\rule{0ex}{0ex}}$

Components of these magnetic inductions, parallel to plane of loop neutralise each other. Hence, resultant of these two magnetic inductions is $2{B}_{0}\mathrm{cos}\theta $ (along the axis).

Similarly, resultantof magnetic inductions produced by currents in remaining two opposite sides BC and AD will also be equal to $2{B}_{0}\mathrm{cos}\theta $ (along the axis in same direction). Hence, resultant magnetic induction.

$B=4{B}_{0}\mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}B=4\times \left(3\times {10}^{-6}\right)\frac{(a/2)}{r}=9\times {10}^{-6}\mathrm{T}=9\mu \mathrm{T}$

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Similar Questions

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 = 1cm^{2}, 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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