KCET · Physics · Magnetic Effects of Current
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
- A \( 6.28 \mathrm{~J} / \mathrm{m} 3 \)
- B \( 0.628 \mathrm{~J} / \mathrm{m} 3 \)
- C \( 3.14 \mathrm{~J} / \mathrm{m} 3 \)
- D \( 0.314 \mathrm{~J} / \mathrm{m} 3 \)
Answer & Solution
Correct Answer
(B) \( 0.628 \mathrm{~J} / \mathrm{m} 3 \)
Step-by-step Solution
Detailed explanation
Magnetic field inside a toroid
\(\mathrm{B}=\mu_{o} \mathrm{nl}\)
Magnetic energy density inside the toroid is
\(u_{o}=\frac{B^{2}}{2 \mu_{o}}=\frac{\mu_{o}^{2} n^{2} I^{2}}{2 \mu_{o}}=\frac{\mu_{o} n^{2} I^{2}}{2}\)
\(=\frac{4 \Pi \times 10^{-7} \times(500)^{2} \times 2^{2}}{2}=0.628 \mathrm{~J} \mathrm{~m}^{-3}\)
\(\mathrm{B}=\mu_{o} \mathrm{nl}\)
Magnetic energy density inside the toroid is
\(u_{o}=\frac{B^{2}}{2 \mu_{o}}=\frac{\mu_{o}^{2} n^{2} I^{2}}{2 \mu_{o}}=\frac{\mu_{o} n^{2} I^{2}}{2}\)
\(=\frac{4 \Pi \times 10^{-7} \times(500)^{2} \times 2^{2}}{2}=0.628 \mathrm{~J} \mathrm{~m}^{-3}\)
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