| A short bar magnet is placed in the magnetic meridian of the earth with North Pole pointing north. Neutral points are found at a distance of 30 cm from the magnet on the East - West line, drawn through the middle point of the magnet. |
| The magnetic moment of the magnet in \[A{{m}^{2}}\]is close to: [JEE MAIN 11-04-2015] |
| (Given \[\frac{{{\mu }_{0}}}{4\pi }={{10}^{-7}}\] in SI units and \[{{B}_{H}}=\]Horizontal component of earth's magnetic field \[=3.6\times {{10}^{-}}^{5}\] Tesla.) |
A) 9.7
B) 4.9
C) 19.4
D) 14.6
Correct Answer: A
Solution :
| [a] At 30cm from the magnet on its equitorial plane \[{{\vec{B}}_{magnet}}=-\overrightarrow{{{B}_{M}}}\](newtral point) |
| so by equating their magnitude \[\frac{{{\mu }_{0}}}{4\pi }\frac{M}{{{r}^{3}}}=3.6\times {{10}^{-5}}\]Tesla |
| \[\frac{{{10}^{-7}}\times M}{{{(0.3)}^{3}}}=3.6\times {{10}^{-5}}\]Tesal |
| \[M=3.6\times 0.027\times {{10}^{2}}=9.7A{{m}^{2}}\] |
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