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A permanent magnet in the shape of a thin cylinder of length $10 \mathrm{~cm}$ has $\mathrm{M}=10^6 \mathrm{~A} / \mathrm{m}$. Calculate the magnetisation current $\mathrm{I}_{\mathrm{M}}$ -
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Verified Answer
As we know that :
$\mathrm{M}$ (Magnetic moment) $=\frac{\mathrm{I}_{\mathrm{m}}}{l}$
As given that, $\mathrm{M}$ (intensity of magnetisation) $=10^6 \mathrm{~A} / \mathrm{m}$. $l($ length $)=10 \mathrm{~cm}=10 \times 10^{-2} \mathrm{~m}=0.1 \mathrm{~m}$
$\mathrm{I}_{\mathrm{m}}$ (magnetisation current) $=$ ?
From(i)
$\mathrm{M}=\frac{\mathrm{I}_{\mathrm{m}}}{l}$
So,
$$
\begin{aligned}
&\mathrm{I}_{\mathrm{M}}=\mathrm{M} \times l=10^6 \times 0.1 \\
&\mathrm{I}_{\mathrm{M}}=10^5 \mathrm{~A}
\end{aligned}
$$
$\mathrm{M}$ (Magnetic moment) $=\frac{\mathrm{I}_{\mathrm{m}}}{l}$
As given that, $\mathrm{M}$ (intensity of magnetisation) $=10^6 \mathrm{~A} / \mathrm{m}$. $l($ length $)=10 \mathrm{~cm}=10 \times 10^{-2} \mathrm{~m}=0.1 \mathrm{~m}$
$\mathrm{I}_{\mathrm{m}}$ (magnetisation current) $=$ ?
From(i)
$\mathrm{M}=\frac{\mathrm{I}_{\mathrm{m}}}{l}$
So,
$$
\begin{aligned}
&\mathrm{I}_{\mathrm{M}}=\mathrm{M} \times l=10^6 \times 0.1 \\
&\mathrm{I}_{\mathrm{M}}=10^5 \mathrm{~A}
\end{aligned}
$$
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