An iron rod is subjected to cycles of magnetisation at the rate of 50Hz. Given the density of the rod is 8 × 103kg/m3 and specific heat is 0.11 × 10–3 cal/ kg0C. The rise in temperature per minute, if the area enclosed by the B – H loop corresponds to energy of 10–2 J is (Assume there is no radiation losses)
n = 50 per second
n per minute = 50 × 60 = 3000
n x area = ms
3000 x 10-2 = 4.2 x 8 x 103 x 0.11 x 10-3