Magnetic field due to current element

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Question

Two circular coils X and Y, having equal number of turns and carrying currents in the same sense, subtend same solid angle at point O. If the smaller coil X is midway between O and Y and if we represent the magnetic induction due to bigger coil Y at O as B_y and that due to smaller coil Y at O as B_x, then find the ratio B_x/B_y.

Moderate

Solution

$B_{y} = \frac{μ_{0}}{4 π} \frac{2 π I (2 r)^{2}}{{[(2 r)^{2} + (2 d)^{2}]}^{3 / 2}}, B_{x} = \frac{μ_{0}}{4 π} \frac{2 π I r^{2}}{{[r^{2} + d^{2}]}^{3 / 2}} solve to get \frac{B_{x}}{B_{y}} = 2$

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Magnetic field due to current element

Unlock the full solution & master the concept.

By=μ04π2πI(2r)2(2r)2+(2d)23/2,Bx=μ04π2πIr2r2+d23/2 solve to get BxBy=2

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$B_{y} = \frac{μ_{0}}{4 π} \frac{2 π I (2 r)^{2}}{{[(2 r)^{2} + (2 d)^{2}]}^{3 / 2}}, B_{x} = \frac{μ_{0}}{4 π} \frac{2 π I r^{2}}{{[r^{2} + d^{2}]}^{3 / 2}} solve to get \frac{B_{x}}{B_{y}} = 2$