A) \[B\alpha \frac{1}{{{x}^{3/2}}}\]
B) \[B\alpha \frac{1}{{{x}^{2}}}\]
C) \[B\alpha \frac{1}{{{x}^{3}}}\]
D) \[B\alpha \frac{1}{{{x}^{1/2}}}\]
Correct Answer: C
Solution :
\[B=\frac{{{\mu }_{0}}NI{{R}^{2}}}{2{{({{R}^{2}}+{{x}^{2}})}^{3/2}}}\] Here, \[x>>R\] \[(\therefore {{R}^{2}}+{{x}^{2}}={{x}^{2}})\] \[\therefore \] \[B=\frac{{{\mu }_{0}}NI{{R}^{2}}}{2{{x}^{3}}}\] Or \[B=\frac{1}{{{x}^{3}}}\]
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