A) \[125\mu T\]
B) \[150\mu T\]
C) \[250\mu T\]
D) \[75\mu T\]
Correct Answer: C
Solution :
[c] \[B=\frac{{{\mu }_{0}}i\,\,{{a}^{2}}}{1{{({{x}^{2}}+{{a}^{2}})}^{3/2}}}\] \[B'=\frac{{{\mu }_{0}}i}{2a}=\frac{{{\mu }_{0}}i\,{{a}^{2}}}{2a{{({{x}^{2}}+{{a}^{2}})}^{3/2}}}\left( \frac{{{({{x}^{2}}+{{a}^{2}})}^{3/2}}}{{{a}^{2}}} \right)\] \[B'=\frac{B.{{({{x}^{2}}+{{a}^{2}})}^{3/2}}}{{{a}^{3}}}\] Put \[x=4\And a=3\Rightarrow B'=\frac{54({{5}^{3}})}{3\times 3\times 3}=250\mu T\]
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