question_answer

A bar magnet is 10 cm long is kept with its north (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)  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 \[{{B}_{H}}\] \[{{B}_{H}}=\frac{{{\mu }_{0}}}{4\pi }.\frac{M}{{{(O{{P}^{2}}+A{{O}^{2}})}^{3/2}}}\] \[0.4\times {{10}^{-4}}={{10}^{-7}}\times \frac{M}{{{(200\times {{10}^{-4}}+25\times {{10}^{-4}})}^{3/2}}}\] \[\frac{0.4\times {{10}^{-4}}}{{{10}^{-7}}}\times {{(225\times {{10}^{-4}})}^{3/2}}=M\] \[0.4\times {{10}^{3}}\times {{10}^{-6}}{{(225)}^{3/2}}=M\] \[M=1.35A.m\]