A) 18 G
B) 19.7 G
C) 15.7 G
D) 17.7G
Correct Answer: C
Solution :
(c) 15.7 G Here, \[N=150,\,R=12\,cm=12\times {{10}^{-3}}m,\,l=2A\] \[\therefore B=\frac{{{\mu }_{0}}Nl}{2R}=\frac{4\pi \times {{10}^{-7}}\times 150\times 2}{2\times 12\times {{10}^{-2}}}\] \[=1.57\times {{10}^{-3}}T\] \[=1.57\times {{10}^{-3}}T=15.7\times {{10}^{-4}}T=15.7\,G\]You need to login to perform this action.
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