The relation between \(\mu\) and \(H\) for a specimen of iron is \(\mu = \left[ \frac{0.4}{H} + 12 \times 10^{-4} \right] \text{Hm}^{-1}\). The value of \(H\) which produces flux density of 1 T will be (\(\mu\) = magnetic permeability, \(H\) = magnetic intensity)
Show Hint
- \( 250 \text{ Am}^{-1} \)
- \( 500 \text{ Am}^{-1} \)
- \( 750 \text{ Am}^{-1} \)
- \( 10^3 \text{ Am}^{-1} \)
The Correct Option is B
Solution and Explanation
Given: \(\mu = \left[ \frac{0.4}{H} + 12 \times 10^{-4} \right] \text{Hm}^{-1}\) Flux density \(B = 1 \text{ T}\) We know that \(B = \mu H\). Substituting the given relation for \(\mu\), we have \[ B = \left[ \frac{0.4}{H} + 12 \times 10^{-4} \right] H \] \[ B = 0.4 + 12 \times 10^{-4} H \] We are given \(B = 1 \text{ T}\), so \[ 1 = 0.4 + 12 \times 10^{-4} H \] \[ 0.6 = 12 \times 10^{-4} H \] \[ H = \frac{0.6}{12 \times 10^{-4}} \] \[ H = \frac{6 \times 10^{-1}}{12 \times 10^{-4}} \] \[ H = \frac{1}{2} \times 10^3 \] \[ H = 500 \text{ Am}^{-1} \] Therefore, the value of \(H\) is \(500 \text{ Am}^{-1}\).
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