Two circular loops having same radius [ R = 1 | vedclass

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Question

(C) The magnetic field on the axis of a circular loop of radius $R$ at a distance $x$ from its center is given by $B = \frac{\mu_{0} I R^{2}}{2(R^{2}

Vedclass · Accepted Answer

$\frac{56 \mu_{0}}{\sqrt{5}} \, T$

Vedclass · Answer

(C) The magnetic field on the axis of a circular loop of radius $R$ at a distance $x$ from its center is given by $B = \frac{\mu_{0} I R^{2}}{2(R^{2} + x^{2})^{3/2}}$.Here,$R = 10 \, cm = 0.1 \, m$,$I = \frac{7}{2} \, A$,and the point $P$ is at the midpoint,so $x = 5 \, cm = 0.05 \, m$ from each coil.Since the currents flow in the same direction,the magnetic fields $B_{1}$ and $B_{2}$ from both coils at point $P$ are in the same direction.$B_{net} = B_{1} + B_{2} = 2 	imes \left( \frac{\mu_{0} I R^{2}}{2(R^{2} + x^{2})^{3/2}} ight) = \frac{\mu_{0} I R^{2}}{(R^{2} + x^{2})^{3/2}}$.Substituting the values:$B_{net} = \frac{\mu_{0} 	imes (7/2) 	imes (0.1)^{2}}{((0.1)^{2} + (0.05)^{2})^{3/2}} = \frac{\mu_{0} 	imes 3.5 	imes 0.01}{(0.01 + 0.0025)^{3/2}} = \frac{0.035 \mu_{0}}{(0.0125)^{3/2}}$.Since $0.0125 = \frac{125}{10000} = \frac{5^{3}}{100^{2}}$,we have $(0.0125)^{3/2} = (\frac{5}{100})^{3} 	imes \sqrt{5} = \frac{125}{10^{6}} \sqrt{5} = 0.000125 \sqrt{5}$.$B_{net} = \frac{0.035 \mu_{0}}{0.000125 \sqrt{5}} = \frac{35000 \mu_{0}}{125 \sqrt{5}} = \frac{280 \mu_{0}}{\sqrt{5}} 	imes \frac{1}{5} 	imes 5 = \frac{56 \mu_{0}}{\sqrt{5}} \, T$.

Two circular loops having same radius $[ R = 10 \, cm ]$ and same current $[ I = \frac{7}{2} \, A ]$ are placed along the same axis as shown. If the distance between their centers is $[ 10 \, cm ]$,find the net magnetic field at point $P$ located at the midpoint between them.

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