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Question #43871

Difficulty: 8/10 | 🟢 Free

The Question

An inductor is formed by winding a coil around a paramagnetic core. Due to the Curie law, the inductance varies with the core's absolute temperature $T$ as $L(T) = \frac{\alpha}{T}$ , where $\alpha$ is a positive constant. The inductor is connected in a closed series circuit with a resistor of resistance $R$ . The resistor is perfectly thermally coupled to the core, meaning all the electrical power dissipated in the resistor is transferred as heat to the core. The core has a constant heat capacity $C_h$ , and no heat is lost to the surroundings. At time $t=0$ , the temperature of the core is $T_0$ and the current in the circuit is $I_0$ . The induced EMF in the inductor is given by Faraday's law as $V_L = -\frac{d}{dt}(LI)$ . Which of the following expressions gives the final temperature $T_f$ of the core as $t \to \infty$ ?

Options

A.
$T_0 + \frac{\alpha I_0^2}{2 C_h T_0}$
B.
$\sqrt{T_0^2 + \frac{\alpha I_0^2}{C_h}}$
C.
$T_0 \exp\left( \frac{\alpha I_0^2}{2 C_h T_0^2} \right)$
D.
$T_0 \exp\left( \frac{\alpha I_0^2}{C_h T_0^2} \right)$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Curie’s Law|Maxwell-Boltzmann Distribution|Transient Current in RL Circuit

Why This Question Is Tricky

Common Mistakes

This question tests the ability to handle a non-linear, coupled electro-thermal system where the inductance is time-varying due to temperature changes.

The primary trap is misapplying the principle of conservation of energy.

Students commonly assume that the total heat dissipated in the resistor equals the initial energy stored in the inductor ( $Q = \frac{1}{2}L_0 I_0^2$ ), leading to Option A.

However, this ignores the thermodynamic work done by the changing inductance (the $I^2 dL$ term in the power expansion).

Another common mistake is treating the induced EMF simply as $L \frac{dI}{dt}$ instead of $\frac{d}{dt}(LI)$ , which ignores the $I \frac{dL}{dt}$ back-EMF contribution and leads to Option B.

The correct approach requires solving the exact differential equation governing the coupled system.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Magnetism and Matter - Class 12
Concept	Curie’s Law
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free