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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
Options:
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
Solution:
1769 Upvotes
Verified Answer
The correct answer is:
$0.6 \mathrm{~m}$
$\mathrm{F}=\mathrm{mB} \quad$ 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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