MHT CET · Physics · Magnetic Properties of Matter
A bar magnet having length \(5 \mathrm{~cm}\) and area of cross-section \(4 \mathrm{~cm}^{2}\) has magnetic moment \(2 \mathrm{Am}^{2}\). If magnetic susceptibility is \(5 \times 10^{-6}\), the magnetic intensity will be
- A \(0 \cdot 2 \times 10^{10} \frac{\mathrm{A}}{\mathrm{m}}\)
- B \(0.5 \times 10^{10} \frac{\mathrm{A}}{\mathrm{m}}\)
- C \(5 \times 10^{10} \frac{\mathrm{A}}{\mathrm{m}}\)
- D \(2 \times 10^{10} \frac{\mathrm{A}}{\mathrm{m}}\)
Answer & Solution
Correct Answer
(D) \(2 \times 10^{10} \frac{\mathrm{A}}{\mathrm{m}}\)
Step-by-step Solution
Detailed explanation
\(\text {Volume of the magnet } \mathrm{V} =\ell \times \mathrm{A} \)
\( =5 \mathrm{~cm} \times 4 \mathrm{~cm}^{2}=20 \mathrm{~cm}^{3} \)
\( =20 \times 10^{-6} \mathrm{~m}^{3}=2 \times 10^{-5} \mathrm{~m}^{3} \)
\( \text {magnetic moment }=2 \mathrm{Am}^{2} \)
\( \therefore \text {Magnetization } \mathrm{M} =\frac{2 \times 10^{-5}}{2}=10^{-5} \mathrm{~A} / \mathrm{m} \)
\( \text {Magnetic intensity } \mathrm{H} =\frac{\mathrm{M}}{\chi}=\frac{10^{-5}}{5 \times 10^{-6}}\) \(=2 \times 10^{10} \mathrm{~A} / \mathrm{m}\)
\( =5 \mathrm{~cm} \times 4 \mathrm{~cm}^{2}=20 \mathrm{~cm}^{3} \)
\( =20 \times 10^{-6} \mathrm{~m}^{3}=2 \times 10^{-5} \mathrm{~m}^{3} \)
\( \text {magnetic moment }=2 \mathrm{Am}^{2} \)
\( \therefore \text {Magnetization } \mathrm{M} =\frac{2 \times 10^{-5}}{2}=10^{-5} \mathrm{~A} / \mathrm{m} \)
\( \text {Magnetic intensity } \mathrm{H} =\frac{\mathrm{M}}{\chi}=\frac{10^{-5}}{5 \times 10^{-6}}\) \(=2 \times 10^{10} \mathrm{~A} / \mathrm{m}\)
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