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JEE Main 2015 (11 Apr, Shift 1) · Earth's Magnetism

Earth's Magnetism question from JEE Main 2015 (11 Apr, Shift 1)

Q12JEE Main 2015 · 11 APR · SHIFT 1Medium

Earth's Magnetism

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 μ04π\frac{\mu_0}{4\pi} = 10−-7 in SI units and BH = Horizontal component of earth's magnetic field = 3.6 ×\times 10 −-5 Tesla.)

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Answer: (a)

  1. N pole pointing north: neutral points lie on the equatorial (east-west) line, where the magnet's field points south and cancels BHB_H.
  2. μ04πMr3=BH⇒M=BHr310−7\frac{\mu_0}{4\pi}\frac{M}{r^3} = B_H \Rightarrow M = \frac{B_H r^3}{10^{-7}}
  3. M=3.6×10−5×(0.3)310−7=3.6×10−5×2.7×105=9.72M = \frac{3.6 \times 10^{-5} \times (0.3)^3}{10^{-7}} = 3.6 \times 10^{-5} \times 2.7 \times 10^{5} = 9.72
  4. So the answer is (a) 9.7 A m2^2.
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