A) \[6.53\times {{10}^{-}}\text{A}\]
B) \[13.25\times {{10}^{-10}}\text{A}\]
C) \[9.6\times {{10}^{6}}\text{A}\]
D) \[1.04\times {{10}^{-3}}\text{A}\]
Correct Answer: D
Solution :
Magnetic field produced at the centre of circular orbit \[B=\frac{{{\mu }_{0}}}{4\pi }\left( \frac{2\pi I}{r} \right)\] or \[I=\frac{4\pi }{{{\mu }_{0}}}\left( \frac{rB}{2\pi } \right)\] Given: \[r=5.2\times {{10}^{-11}}m,\] \[B=12.56T\] So, \[I=\frac{{{10}^{7}}\times 5.2\times {{10}^{-11}}\times 12.56}{2\times 3.14}\] \[=1.04\times {{10}^{-3}}A\]
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