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An iron rod is subjected to cycles of magnetization at the rate of 50 Hz. Given density of rod is 8 x 10³ kg/m³ and specific heat 0.11 x 10^-3 cal/kg°C. If area enclosed by the B-H loop corresponds to energy of 10^-2 J then rise in temperature per minute is nearly is (Assume there is no radiation losses)

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An Intiative by Sri Chaitanya

78.5°C

15°C

8°C

0.13°C

answer is C.

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Detailed Solution

$E = (n t A) V$

$m s Δ t = n t A V$

$Δ t = \frac{n t A}{S d } = 8 ° C$

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