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JEE Main 2015 · 11 Apr · Shift 1, Physics question 21

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 Am2^2 is close to :

(Given μ04π=10−7\dfrac{\mu_0}{4\pi}=10^{-7} in SI units and BHB_H = Horizontal component of earth's magnetic field = 3.6×10−53.6\times10^{-5} Tesla.)

Choose one option
Answer and worked solution

Official answer: A

On the equatorial (E-W) line the magnet's field opposes BHB_H, giving neutral points where
μ04πMd3=BH  ⟹  M=3.6×10−5×(0.3)310−7≈9.7 A m2.\frac{\mu_0}{4\pi}\frac M{d^3}=B_H\implies M=\frac{3.6\times10^{-5}\times(0.3)^3}{10^{-7}}\approx9.7\text{ A m}^2.

Topics:Earth's magnetismMagnetic dipole and bar magnet
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