A) 40 A
B) 4 A
C) 0.4 A
D) 0.2 A
Correct Answer: C
Solution :
\[r=5cm.=5\times {{10}^{-2}}m\]\[{{B}_{E}}=0.5\times {{10}^{-5}}W/{{m}^{2}}\]we know that field due to coil at centre \[B=\frac{{{\mu }_{0}}I}{2r}\] it annuals the earth's megnetic field So, \[\frac{{{\mu }_{0}}I}{2r}=0.5\times {{10}^{-5}}\] \[I=\frac{2R\times 0.5\times {{10}^{-5}}}{\mu }=\frac{5}{4\pi }A=0.4A\]You need to login to perform this action.
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