Question

An iron rod is subjected to cycles of magnetization at the rate of $50Hz$. Given the density of the rod is $8 \times {10^3}kg/{m^3}$ and specific heat is $0.11 \times {10^{ - 3}}Cal/kg^\circ C$. 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 are no radiation losses).
A. $78^\circ C$
B. $88^\circ C$
C. $8.1^\circ C$
D. None of these

Answer 1

To solve this problem, we need to use the formula of heat produced per minute which is related to mass of the rod, specific heat and the temperature rise. As the energy and frequency are given, we can find the heat produced first. Finally, we will determine the temperature rise per minute.

Formula used:
$Q = mc\Delta T$,
where, $Q$ is the heat produced, $m$ is the mass, $c$ is the specific heat and $\Delta T$ is the rise in the temperature.

Complete step by step answer:
Here we are given that An iron rod is subjected to cycles of magnetization at the rate of $50Hz$ and the area enclosed by the B-H loop corresponds to energy of ${10^{ - 2}}J$.
Therefore, the heat produced per unit volume is:

\Rightarrow\dfrac{Q}{V} = \dfrac{{30}}{{4.2}}J$$ We also know that,

Q = mc\Delta T \\
\Rightarrow \dfrac{Q}{V} = \dfrac{{mc\Delta T}}{V} = \dfrac{{\rho c\Delta T}}{V} \\
\Rightarrow \Delta T = \dfrac{{{{10}^{ - 2}} \times 50 \times 60}}{{4.2 \times 8 \times {{10}^3} \times 0.11 \times {{10}^{ - 3}}}} \\
\therefore \Delta T = 8.1^\circ C

Thus, the temperature rise per minute is $8.1^\circ C$. **Hence, option C is the right answer.** **Note:** In this question, we have determined the rise in temperature by equating energy loss per unit volume per unit cycle with area of hysteresis loop Energy loss per second per unit volume. In this problem, we need to note that the heat produced is per unit volume. This is because we are given the density of an iron rod and not the mass of it.