A) \[5.65\times {{10}^{-5}}T\]
B) \[5.27\times {{10}^{-5}}T\]
C) \[6.54\times {{10}^{-5}}T\]
D) \[9.20\times {{10}^{-5}}T\]
Correct Answer: A
Solution :
(a) \[5.65\times {{10}^{-5}}T\] \[B=\frac{{{\mu }_{0}}I{{R}^{2}}}{2{{({{R}^{2}}+{{x}^{2}})}^{3/2}}}\] Here, \[l=12.5\,A,\,R=3\,cm=3\times {{10}^{-2}}m\] \[x=4\,cm=4\times {{10}^{-2}}m\] \[\therefore B=\frac{4\pi \times {{10}^{-7}}\times 12.5\times {{(3\times {{10}^{-2}})}^{2}}}{2{{[{{(3\times {{10}^{-2}})}^{2}}+{{(4\times {{10}^{-2}})}^{2}}]}^{3/2}}}=5.65\times {{10}^{-5}}T.\]
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