MHT CET · Physics · Magnetic Effects of Current
A bar magnet of magnetic moment \(5 \mathrm{Am}^{2}\) is placed in a uniform magnetic induction \(3 \times 10^{-5} \mathrm{~T}\). If each pole of a magnet experiences a force of \(2 \cdot 5 \times 10^{-4} \mathrm{~N}\) then the magnetic length of the magnet is
- A \(0.8 \mathrm{~m}\)
- B \(0 \cdot 2 \mathrm{~m}\)
- C \(0.6 \mathrm{~m}\)
- D \(0 \cdot 4 \mathrm{~m}\)
Answer & Solution
Correct Answer
(C) \(0.6 \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{F}=\mathrm{mB} \quad\) where \(\mathrm{m}\) is pole strength.
\(
\therefore \mathrm{m}=\frac{\mathrm{F}}{\mathrm{B}}=\frac{2.5 \times 10^{-4}}{3 \times 10^{-5}}=\frac{25}{3} \mathrm{Am}
\)
Magnetic moment \(\mathrm{M}=\mathrm{mL}\)
\(
\therefore \mathrm{L}=\frac{\mathrm{M}}{\mathrm{m}}=\frac{5 \times 3}{25}=0.6 \mathrm{~m}
\)
\(
\therefore \mathrm{m}=\frac{\mathrm{F}}{\mathrm{B}}=\frac{2.5 \times 10^{-4}}{3 \times 10^{-5}}=\frac{25}{3} \mathrm{Am}
\)
Magnetic moment \(\mathrm{M}=\mathrm{mL}\)
\(
\therefore \mathrm{L}=\frac{\mathrm{M}}{\mathrm{m}}=\frac{5 \times 3}{25}=0.6 \mathrm{~m}
\)
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