A bar magnet of length $10$ cm lies in the magnetic meridian with its north pole pointing north. Find the distance between the neutral points if the horizontal component of the earth's magnetic field is $0.38 \times 10^{- 4}$ wb $/ m^{2}$ and the magnetic moment is $0.747 A m^{2}$ .

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Solution

At a null point, magnet's magnetic field is balanced by the earth's field.
as angle of dip $= 0^{\circ}$
$B_{m a g n e t} = \frac{μ_{0}}{4 π} \frac{M}{r^{3}}$
$B_{H} = B_{m a g n e t}$
$0.3 * 10^{- 4} = 10^{- 7} * \frac{0.747}{r^{3}}$
$∴ r^{3} = 1.93 * 10^{- 3}$
$∴ r = 0.125 m$
$r = 12.5 c m$
$∴$ Distance between neutral points $= 2 r = 25 c m$

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A bar magnet of length 10cm lies in the magnetic meridian with its north pole pointing north. Find the distance between the neutral points if the horizontal component of the earth's magnetic field is 0.38×10−4 wb/m2 and the magnetic moment is 0.747Am2.

At a null point, magnet's magnetic field is balanced by the earth's field.as angle of dip =0∘ Bmagnet=μ04πMr3 BH=Bmagnet 0.3∗10−4=10−7∗0.747r3 ∴r3=1.93∗10−3 ∴r=0.125m r=12.5cm∴ Distance between neutral points =2r=25cm

A bar magnet of length $10$ cm lies in the magnetic meridian with its north pole pointing north. Find the distance between the neutral points if the horizontal component of the earth's magnetic field is $0.38 \times 10^{- 4}$ wb $/ m^{2}$ and the magnetic moment is $0.747 A m^{2}$ .