A) \[5\,ab-amp\times cm\]
B) \[10\,ab-amp\times cm\]
C) \[2.5\,ab-amp\times cm\]
D) \[20\,ab-amp\times cm\]
Correct Answer: A
Solution :
[a] Here, \[2\ell =8\,cm,\] \[\ell =4cm,\] \[d=\frac{6}{2}=3\,cm\]. At neutral point, \[H=B=\frac{{{\mu }_{0}}}{4\pi }\frac{M}{{{({{d}^{2}}+{{\ell }^{2}})}^{3/2}}}\] \[={{10}^{-7}}\frac{M}{{{(5\times {{10}^{-2}})}^{3}}}=\frac{M}{1250}\] \[\therefore \,\,\,M=1250H=1250\times 3.2\times {{10}^{-5}}\,A{{m}^{2}}\] \[m=\frac{M}{2\ell }=\frac{1250\times 3.2\times {{10}^{-5}}}{8\times {{10}^{-2}}}Am.\,\,\,\,=0.5\] \[Am=0.5\times \frac{1}{10}ab\,\] \[amp\times 100\,cm\] = 5 ab-amp cm.You need to login to perform this action.
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