question_answer

Two non-touching conducting circular loops of radii \[{{R}_{1}}\] and \[{{R}_{2}}\] are concentric. If \[{{R}_{2}}<<{{R}_{1}}\] then the mutual inductance M will be

A)  \[\frac{{{\mu }_{0}}\pi {{R}_{2}}}{R}\]                   

B)  \[\frac{{{\mu }_{0}}{{R}_{2}}}{\pi {{R}_{1}}}\]

C)  \[\sqrt{2}{{\mu }_{0}}\,\pi R_{1}^{2}R_{2}^{2}\]              

D)  \[\frac{{{\mu }_{0}}\pi R_{2}^{2}}{2{{R}_{1}}}\]

E)  \[2\frac{{{R}_{1}}{{R}_{2}}}{{{\mu }_{0}}\,\pi }\]              

Correct Answer: D

Solution :

                Let the current through the outer loop is \[I\]. The magnetic field at the common centre O is \[B=\frac{{{\mu }_{0}}I}{2{{R}_{1}}}\] As,\[{{R}_{2}}<<{{R}_{1}},\]the flux of magnetic field through it will be approximately \[\phi =\frac{{{\mu }_{0}}I}{2{{R}_{1}}}\pi R_{2}^{2}\] Now, mutual inductance,                 \[M=\frac{\phi }{I}=\frac{{{\mu }_{0}}\pi R_{2}^{2}}{2{{R}_{1}}}\]