A) 9 A-m
B) 6.75 A-m
C) 27 A-m
D) 1.35 A-m
Correct Answer: D
Solution :
Length of magnet \[=10cm=10\times {{10}^{-2}}m,\] \[r=15\times {{10}^{-2}}m\] \[OP=\sqrt{225-25}=\sqrt{200}cm\] Since, at the neutral point, magnetic field due to the magnet is equal to \[{{B}_{H}}\] \[{{B}_{H}}=\frac{{{\mu }_{0}}}{4\pi }.\frac{M}{{{(O{{P}^{2}}+A{{O}^{2}})}^{3/2}}}\] \[0.4\times {{10}^{-4}}={{10}^{-7}}\times \frac{M}{{{(200\times {{10}^{-4}}+25\times {{10}^{-4}})}^{3/2}}}\] \[\frac{0.4\times {{10}^{-4}}}{{{10}^{-7}}}\times {{(225\times {{10}^{-4}})}^{3/2}}=M\] \[0.4\times {{10}^{3}}\times {{10}^{-6}}{{(225)}^{3/2}}=M\] \[M=1.35A.m\]You need to login to perform this action.
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