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A long solenoid with \( 40 \) turns per cm carries a current of \( 1 \mathrm{~A} \). The magnetic energy stored per
unit volume is
\( J m^{-3} \).
Options:
unit volume is
\( J m^{-3} \).
Solution:
1480 Upvotes
Verified Answer
The correct answer is:
\( 3.2 \pi \)
Given, number of turns \( =40 \) turns per \( \mathrm{cm} \); current \( =1 \mathrm{~A} \)
Therefore, magnetic energy stored per unit volume \( =\frac{1}{2} \mu_{0} n^{2} I^{2} \)
\[
\begin{array}{l}
=\frac{1}{2} \times\left(4 \Pi \times 10^{-7}\right) \times\left(40 \times 10^{2}\right)^{2} \times(1)^{2} \\
=\frac{4 \times 4 \times 4 \times_{\Pi} \times 10^{-7} \times 10^{5}}{2} \\
=32 \times \Pi \times 10^{-1}=3.2 \Pi
\end{array}
\]
Thus, magnetic energy stored per unit volume \( =3.2 \mathrm{IJ} \mathrm{m}^{-3} \)
Therefore, magnetic energy stored per unit volume \( =\frac{1}{2} \mu_{0} n^{2} I^{2} \)
\[
\begin{array}{l}
=\frac{1}{2} \times\left(4 \Pi \times 10^{-7}\right) \times\left(40 \times 10^{2}\right)^{2} \times(1)^{2} \\
=\frac{4 \times 4 \times 4 \times_{\Pi} \times 10^{-7} \times 10^{5}}{2} \\
=32 \times \Pi \times 10^{-1}=3.2 \Pi
\end{array}
\]
Thus, magnetic energy stored per unit volume \( =3.2 \mathrm{IJ} \mathrm{m}^{-3} \)
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