A) 25.7 Am
B) 23.7 Am
C) 28.7 Am
D) 26.7 Am
Correct Answer: D
Solution :
[d] \[\therefore \,\,\,\,\,\,M=\frac{4\pi }{{{\mu }_{0}}}\frac{{{B}_{H}}{{({{r}^{2}}-{{\ell }^{2}})}^{2}}}{2r}\] \[={{10}^{7}}\times \frac{(0.34\times {{10}^{-4}}).{{[{{(0.40)}^{2}}-{{(0.15)}^{2}}]}^{2}}}{2\times 0.40}\] \[=8.0\,A{{m}^{2}}\] The pole strength of the magnet is, \[m=\frac{M}{2\ell }=\frac{8.0}{0.30}=26.7\,Am\]You need to login to perform this action.
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