A) 6.3 T
B) 12.6 T
C) 18.9 T
D) 25.2 T
Correct Answer: B
Solution :
\[{{B}_{contre}}=\frac{{{\mu }_{0}}I}{2r}\] but \[I=ne\] \[\therefore \] \[{{B}_{centre}}=\frac{{{\mu }_{0}}ne}{2r}\] \[\Rightarrow \]\[{{B}_{centre}}=\frac{4\pi \times {{10}^{-7}}\times 6.6\times {{10}^{15}}\times 1.6\times {{10}^{-19}}}{2\times 0.53\times {{10}^{-10}}}\] \[=12.6\,T\]You need to login to perform this action.
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