A) \[10\mu T\]
B) \[4\mu T\]
C) \[5\mu T\]
D) \[3\mu T\]
Correct Answer: A
Solution :
[a] Current in the element \[=J(2\pi r.dr)\] Current enclosed by ampere loop of radius \[a/2\] \[I=\int\limits_{0}^{a/2}{\frac{{{J}_{0}}r}{a}.2\pi }r.dr\] \[=\frac{2\pi {{J}_{0}}}{3a}{{\left( \frac{a}{2} \right)}^{3}}=\frac{\pi {{J}_{0}}{{a}^{3}}}{12}\] Applying ampere's law \[B.2\pi .\frac{a}{2}={{\mu }_{0}}.\frac{\pi {{J}_{0}}{{a}^{2}}}{12}\Rightarrow B=\frac{{{\mu }_{0}}{{J}_{0}}a}{12}\] On putting values, \[B=10\mu T\]You need to login to perform this action.
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