A) 8 A
B) 6 A
C) 4.5 A
D) 2 A
Correct Answer: A
Solution :
From relation \[B=\mu {{\,}_{0}}H\] or \[\mu {{\,}_{0}}ni=\mu {{\,}_{0}}H\] or \[\frac{Ni}{L}=H\] \[\left( \because \,\,n=\frac{N}{L} \right)\] or \[i=\frac{HL}{N}\] Given, \[H=4\times {{10}^{3}}A{{m}^{-1}}\] \[L=12\,cm=12\times {{10}^{-2}}m\] \[N=60\]turns \[\therefore \] \[i=\frac{4\times {{10}^{3}}\times 12\times {{10}^{-2}}}{60}=8\,A\]You need to login to perform this action.
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