A) \[0.125\,Wb/{{m}^{2}}\]
B) \[1.25\,Wb/{{m}^{2}}\]
C) \[12.5\,Wb/{{m}^{2}}\]
D) \[125\,\,Wb/{{m}^{2}}\]
Correct Answer: C
Solution :
The magnetic field at the centre of the orbit \[B=\frac{{{\mu }_{0}}}{4\pi }\cdot \frac{2\pi (qv)}{r}\] \[={{10}^{-7}}\times \frac{2\times 3.14\times (1.6\times {{10}^{-19}}\times 6.6\times {{10}^{15}})}{0.53\times {{10}^{-10}}}\] \[=12.5\,\,Wb/{{m}^{3}}\]You need to login to perform this action.
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