JEE Mains · Physics · STD 12 - 5. Magnetism and matter
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 \(Am^2\) is close to: (Given \(\frac{{{\mu _0}}}{{4\pi }}\, = 10^{- 7}\) in \(SI\,units\) and \(B_H\, =\) Horizontal component field \(= 3.6\times10^{-5}\, tesla\))
- A \(14.6\)
- B \(19.4\)
- C \(9.7\)
- D \(4.9\)
Answer & Solution
Correct Answer
(C) \(9.7\)
Step-by-step Solution
Detailed explanation
Here, \(r=30\, \mathrm{cm}=0.3 \,\mathrm{cm}\) we know \(\frac{\mu_{0} M}{4 \pi r^{3}}=B_{H}=3.6 \times 10^{-5}\) \(M = \frac{{3.6 \times {{10}^{ - 5}}}}{{{{10}^{ - 7}}}} \times {(0.3)^3}\) Hence, \(M=9.7\, \mathrm{Am}^{2}\)
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