A) 0.28 A
B) 5.57 A
C) 2.8 A
D) None of these
Correct Answer: B
Solution :
Magnetic field at the centre of coil of radius a carrying current i is \[F=\frac{G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}\] \[{{m}_{1}}{{m}_{2}}\] \[\therefore \] Given, \[{{F}_{1}}={{F}_{2}}\] \[\frac{G\left( \frac{M}{81} \right)M}{{{x}^{2}}}=\frac{GM\times M}{{{(60R-x)}^{2}}}\] \[\Rightarrow \] \[\frac{1}{81{{x}^{2}}}=\frac{1}{{{(60R-x)}^{2}}}\]
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