MHT CET · Physics · Magnetic Effects of Current
A cylindrical magnetic rod has length \(5 \mathrm{~cm}\) and diameter \(1 \mathrm{~cm}\). It has uniform magnetization \(5.3 \times 10^3 \mathrm{~A} / \mathrm{m}^3\). Its net magnetic dipole moment is nearly
- A \(1 \times 10^{-2} \mathrm{~J} / \mathrm{T}\)
- B \(0.5 \times 10^{-2} \mathrm{~J} / \mathrm{T}\)
- C \(2.5 \times 10^{-2} \mathrm{~J} / \mathrm{T}\)
- D \(2.08 \times 10^{-2} \mathrm{~J} / \mathrm{T}\)
Answer & Solution
Correct Answer
(D) \(2.08 \times 10^{-2} \mathrm{~J} / \mathrm{T}\)
Step-by-step Solution
Detailed explanation
\(\text {Magnetisation, } M=\frac{\mathrm{m}_{\text {net }}}{\mathrm{V}}\)
\(\therefore \mathrm{m}_{\text {net }}=\mathrm{M} \times \mathrm{V}=\mathrm{M} \times\left(\pi \mathrm{r}^2 l\right)=\mathrm{M} \times \pi ~\times\) \(\frac{\mathrm{d}^2}{4} \times l\)
\(=5.3 \times 10^3 \times 3.142 \times\left(\frac{1 \times 10^{-2}}{2}\right)^2 \times 5\) \(\times~ 10^{-2}\)
\(=2.08 \times 10^{-2} \mathrm{~J} / \mathrm{T}\)
\(\therefore \mathrm{m}_{\text {net }}=\mathrm{M} \times \mathrm{V}=\mathrm{M} \times\left(\pi \mathrm{r}^2 l\right)=\mathrm{M} \times \pi ~\times\) \(\frac{\mathrm{d}^2}{4} \times l\)
\(=5.3 \times 10^3 \times 3.142 \times\left(\frac{1 \times 10^{-2}}{2}\right)^2 \times 5\) \(\times~ 10^{-2}\)
\(=2.08 \times 10^{-2} \mathrm{~J} / \mathrm{T}\)
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