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 is close to: (Given (μ0/4π)=10-7 in SI units and BH = Horizontal component of earth's magnetic field = 3.6 × 10-5 Tesla.)

Question

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 is close to: (Given (μ0/4π)=10-7 in SI units and BH = Horizontal component of earth's magnetic field = 3.6 × 10-5 Tesla.) (A) 9.7 (B) 4.9 (C) 19.4 (D) 14.6

Tardigrade Admin · Accepted Answer

| barB| at an equatorial point =(μo/4 π) (| barm|/r3)=B H ⇒ | barm|=(B H r3/10-7) =(3.6 × 10-5 ×(0.3)3/10-7)= 9.7

9.7

4.9

19.4

14.6

∣Bˉ∣ at an equatorial point =4πμo​​r3∣mˉ∣​=BH​ ⇒∣mˉ∣=10−7BH​r3​ =10−73.6×10−5×(0.3)3​=9.7

$∣ \overset{ˉ}{B} ∣$ at an equatorial point
$= \frac{μ _{o}}{4 π} \frac{∣ m ˉ ∣}{r ^{3}} = B_{H}$
$\Rightarrow ∣ \overset{m}{ˉ} ∣ = \frac{B _{H} r ^{3}}{1 0 ^{- 7}}$
$= \frac{3.6 \times 1 0 ^{- 5} \times ( 0.3 ) ^{3}}{1 0 ^{- 7}} = 9.7$