A) \[\frac{{{R}_{1}}}{{{R}_{2}}}\]
B) \[\frac{{{R}_{2}}}{{{R}_{1}}}\]
C) \[\frac{R_{1}^{2}}{{{R}_{2}}}\]
D) \[\frac{R_{2}^{2}}{{{R}_{1}}}\]
Correct Answer: D
Solution :
(d) \[\frac{R_{2}^{2}}{{{R}_{1}}}\] Let a current \[{{l}_{1}}\] flows through the outer circular coil of radius \[{{R}_{1}}\]. The magnetic field at the centre of the coil is \[{{B}_{1}}=\frac{{{\mu }_{0}}{{l}_{1}}}{2{{R}_{1}}}\] As the inner coil of radius \[{{R}_{2}}\] placed co-axially has small radius \[({{R}_{2}}<{{R}_{1}}),\] therefore \[{{B}_{1}}\] may be taken constant over its cross-sectional area. Hence, flux associated with the inner coil is \[{{\phi }_{1}}={{B}_{1}}\pi R_{2}^{2}=\frac{{{\mu }_{0}}{{l}_{1}}}{2{{R}_{1}}}\pi R_{2}^{2}\] As \[M=\frac{{{\phi }_{2}}}{{{l}_{1}}}=\frac{{{\mu }_{0}}\pi R_{2}^{2}.}{2{{R}_{1}}};\therefore \,\,M\propto \frac{R_{2}^{2}}{{{R}_{1}}}\]
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