Two concentric and coplanar rings of radius r | vedclass

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(B) The magnetic field produced by the larger ring of radius $R$ at its center is $B_R = \frac{\mu_0 I}{2R}$.Since $r < R$,we can assume this magnetic

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$\frac{\pi }{2R}{\mu _0}Ir\left( {R - r} \right)$

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(B) The magnetic field produced by the larger ring of radius $R$ at its center is $B_R = \frac{\mu_0 I}{2R}$.Since $r The magnetic field produced by the smaller ring of radius $r$ at its center is $B_r = \frac{\mu_0 I}{2r}$.Since the currents flow in opposite directions,the net magnetic field at the center is $B = |B_r - B_R| = \frac{\mu_0 I}{2} \left( \frac{1}{r} - \frac{1}{R} ight)$.The magnetic flux $\phi$ linked through the smaller coil is given by $\phi = B \cdot A$,where $A = \pi r^2$ is the area of the smaller coil.Substituting the values,we get $\phi = \left[ \frac{\mu_0 I}{2} \left( \frac{R - r}{rR} ight) ight] 	imes \pi r^2$.Simplifying this,we get $\phi = \frac{\pi \mu_0 I r (R - r)}{2R}$.

Two concentric and coplanar rings of radius $r$ and $R$ are shown in the figure. Find the magnetic flux linked through the smaller coil when the same current $I$ flows through each in opposite directions $(r << R)$.

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