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 ${Am}^{2}$ is close to:

(Given $\frac{μ_{0}}{4 π} = 10^{- 7}$ in SI units and $B_{H} =$ Horizontal component of earth's magnetic field $= 3.6 \times 10^{- 5}$ Tesla.)

Question

Aicharya · Accepted Answer

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 \frac{μ_{0}}{4 π} \frac{M}{x^{3}} = B_{H}

\Rightarrow \frac{10^{- 7} \times M}{{(\frac{30}{100})}^{3}} = 3.6 \times 10^{- 5}

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

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

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

= 9.72

M \approx 9.7

A short bar magnet is placed in the magnetic meridian of the earth with North Pole pointing north. Neutral…

Solution