A)
B)
C)
D)
Correct Answer: A
Solution :
\[dB=\frac{{{\mu }_{0}}(dq)}{2r}\left( \frac{\omega }{2\pi } \right)\] \[B=\int{dB=\frac{{{\mu }_{0}}\omega }{4\pi }.\frac{Q}{\pi {{R}^{2}}}2\pi \int\limits_{0}^{R}{\frac{rdr}{r}}}\] \[B=\frac{{{\mu }_{0}}\omega Q}{2\pi {{R}^{2}}}.\,R\] \[B=\frac{{{\mu }_{0}}\omega Q}{2\pi R}\] \[B\propto \frac{1}{R}\]
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