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JEE PHYSICS MCQ: MAGNETISM AND MATTER

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The relation between μ\mu and HH for a specimen of iron is given by μ=(0.8H+8×10−4)H/m.\mu = \left( {\frac{{0.8}}{H} + 8 \times {{10}^{ - 4}}} \right)H/m. The value of HH which produces flux density of 1 tesla will be
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Correct option: A
250 A/m
Option A (250 A/m) is correct because it is calculated by substituting the given values into the formula B = μH and solving for H.

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