Question 23 of 57

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 $30cm$ from the magnet on the East-West line, drawn through the middle point of the magnet. The magnetic moment of the magnet in ${Am}^{2}$ is close to:

(Given $\dfrac{{\mu }_{0}}{4\pi }={10}^{-7}$ in SI units and ${B}_{H}=$ Horizontal component of earth's magnetic field $=3.6\times {10}^{-5}$ Tesla.)

A4.9B14.6C19.4D9.7

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Answer: (D) 9.7

Horizontal component of earth's magnetic field is along geographical south its North.

$\vec{M}=$ Magnetic dipole moment.



At neutral point on the equatorial line the magnetic field ${B}^{'}$ cancels ${B}_{H}$

$∴{B}^{'}={B}_{H}$

$\Rightarrow \dfrac{{\mu }_{0}}{4\pi }\dfrac{M}{{x}^{3}}={B}_{H}$

$\Rightarrow \dfrac{{10}^{-7}\times M}{{\left(\dfrac{30}{100}\right)}^{3}}=3.6\times {10}^{-5}$

$\Rightarrow \dfrac{{10}^{-7}\times {10}^{3}\times M}{27}=3.6\times {10}^{-5}$

$M=\dfrac{27\times 3.6\times {10}^{-5}}{{10}^{-4}}$

$M=27\times 3.6\times {10}^{-1}$

$=9.72$

$M\approx 9.7$