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A toroid has \( 500 \) turns per metre length. If it carries a current of \( 2 \mathrm{~A} \), the magnetic energy
density inside the toroid is
Options:
density inside the toroid is
Solution:
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Verified Answer
The correct answer is:
\( 0.628 \mathrm{~J} / \mathrm{m} 3 \)
(B)Sol. Magnetic field inside a toroid
\[
\mathrm{B}=\mu_{o} \mathrm{nl}
\]
Magnetic energy density inside the toroid is
\[
\begin{array}{l}
u_{o}=\frac{B^{2}}{2 \mu_{o}}=\frac{\mu_{o}^{2} n^{2} I^{2}}{2 \mu_{o}}=\frac{\mu_{o} n^{2} I^{2}}{2} \\
=\frac{4 \Pi \times 10^{-7} \times(500)^{2} \times 2^{2}}{2}=0.628 \mathrm{~J} \mathrm{~m}^{-3}
\end{array}
\]
\[
\mathrm{B}=\mu_{o} \mathrm{nl}
\]
Magnetic energy density inside the toroid is
\[
\begin{array}{l}
u_{o}=\frac{B^{2}}{2 \mu_{o}}=\frac{\mu_{o}^{2} n^{2} I^{2}}{2 \mu_{o}}=\frac{\mu_{o} n^{2} I^{2}}{2} \\
=\frac{4 \Pi \times 10^{-7} \times(500)^{2} \times 2^{2}}{2}=0.628 \mathrm{~J} \mathrm{~m}^{-3}
\end{array}
\]
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