A) 9 A-m
B) 6.75 A-m
C) 27 A-m
D) 13.5 A-m
Correct Answer: D
Solution :
Since, \[H=\frac{{{\mu }_{0}}}{4\pi }.\frac{M}{{{(15\times {{10}^{-2}})}^{3}}}\] \[\therefore \] \[(M=m\times 2l)\] or \[0.4\times {{10}^{-4}}={{10}^{-7}}\times \frac{m\times 10\times {{10}^{-2}}}{15\times 15\times 15\times {{10}^{-6}}}\] or \[\frac{0.4\times {{10}^{-4}}\times 15\times 15\times 15\times {{10}^{-6}}}{{{10}^{-8}}}=m\] or \[m=1350\times {{10}^{-2}}\] or \[m=13.5A-m\] Hence, pole strength \[m=13.5\,A-m\]You need to login to perform this action.
You will be redirected in
3 sec