question_answer

A bar magnet of 10 cm long is kept with its N-pole pointing north. A neutral point is formed at a distance of 15 cm from each pole. Given the horizontal component of earths field is 0.4 gauss, the pole strength of the magnet is

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 \[m=13.5\,A-m\]