Question

The magnetic dipole moment of a small bar magnet is $1.6A{m}^2$. It is placed in magnetic meridian with its north pole pointing toward geographic south of the earth. If the neutral point is obtained at $20$ cm from the centre of the magnet then, the horizontal component of earth's magnetic field is:

• A. ${10}^{-5}T$
• B. $4\times {10}^{-5}T$
• C. $4\times {10}^{+5}T$
• D. $4\times {10}^{-7}T$

Answer 1

The correct option is B $4\times {10}^{-5}T$

at the neutral point, the net magnetic field in zero.

The field on a axial line (Bax) due to the magnetic

dipole is given by at a distance $r=0.2m$

$Bax=\frac{{\mu }_0}{4\pi }\frac{2m}{r^3}$ (M : Magnetic moment =$1.6A{m}^2$)

$\Rightarrow Bax=\frac{{10}^{-7}\times 2\times 1.6}{(0.2{)}^3}=4\times {10}^{-5}T$

at the neutral point,

$Bax=B_H$

$\Rightarrow B_H=4\times {10}^{-5}T$