A) \[4.2\times {{10}^{-2}}\,weber/{{m}^{2}}\]
B) \[4.2\times {{10}^{-3}}\,weber/{{m}^{2}}\]
C) \[4.2\times {{10}^{-4}}\,weber/{{m}^{2}}\]
D) \[4.2\times {{10}^{-5}}\,weber/{{m}^{2}}\]
Correct Answer: A
Solution :
Field at a point \[x\]from the centre of a current carrying loop on the axis is \[\text{B}=\frac{{{\mu }_{\text{0}}}}{4\pi }.\frac{2M}{{{x}^{3}}}=\frac{{{10}^{-7}}\times 2\times 2.1\times {{10}^{-25}}}{{{({{10}^{-10}})}^{3}}}\] \[\]\[=4.2\times {{10}^{-32}}\times {{10}^{30}}=4.2\times {{10}^{-2}}W/{{m}^{2}}\]
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