A) 2.25m
B) 0.25m
C) 13.0m
D) 1.25m
E) 3.25m
Correct Answer: D
Solution :
Given, \[q=2\mu C=2\times {{10}^{-6}}C\] \[f=6.25\times {{10}^{12}}Hz\] \[B=6.28\text{ }T\] The magnetic field at the centre of the circle \[B=\frac{{{\mu }_{0}}i}{2r}=\frac{{{\mu }_{0}}qf}{2r}\] \[[\because i=qf]\] \[6.28=\frac{4\pi \times {{10}^{-7}}\times 2\times {{10}^{-6}}\times 6.25\times {{10}^{12}}}{2r}\] or\[r=\frac{4\pi \times {{10}^{-7}}\times 2\times {{10}^{-6}}\times 6.25\times {{10}^{12}}}{6.28\times 2}\] \[=1.25m\]
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