JEE Mains · Physics · STD 12 - Magnetism and matter
- JEE Mains 2021
- Hard
- MCQ
A bar magnet of length \(14 \,cm\) is placed in the magnetic meridian with its north pole pointing towards the geographic north pole. A neutral point is obtained at a distance of \(18\, cm\) from the center of the magnet. If \(B _{ H }=0.4 \,G ,\) the magnetic moment of the magnet is \(\left(1\, G =10^{-4} T \right)\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(2.880\, J\, T ^{-1}\)
Step-by-step Solution
Detailed explanation
i.e. \(\frac{2 \mu_{0}}{4 \pi} \frac{ m }{ r ^{2}} \times \frac{7}{ r }=0.4 \times 10^{-4}\)
\(\Rightarrow 2 \times 10^{-7} \times \frac{ m \times 7}{\left(7^{2}+18^{2}\right)^{3 / 2}} \times 10^{4}\)
\(=0.4 \times 10^{-4}\)
\(m =\frac{4 \times 10^{-2} \times(373)^{3 / 2}}{14}\)
\(M = m \times 14 cm = m \times \frac{14}{100}\)
\(=\frac{0.04 \times(373)^{3 / 2}}{14} \times \frac{14}{100}\)
\(=4 \times 10^{-4} \times 7203.82=2.88 J / T\)
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