A thermocol vessel contains 0.5 kg of distilled water at 30°C. A metal coil of area 5 x 10^-3 m^2, n

Q: Why does the current reverse direction?

According to Lenz's Law, the induced current always opposes the change in magnetic flux. When the magnetic field increases, the current flows in one direction to create an opposing field. When the field decreases, the current reverses to try and maintain the field.

Q: Why is the power dissipated constant?

Power dissipated as Joule heating is given by $P = i^2 R$. Since the current is squared, the negative sign during the second half of the cycle becomes positive. The magnitude of the current is constant, so the power is also constant.

Q: Do we need to consider the inductance of the coil?

The problem explicitly states to 'Neglect the inductance of coil.' If we didn't, the current wouldn't be a perfect square wave; it would have exponential growth and decay phases.

Q: Why do we add the heat capacities of the water and the coil?

Both the water and the metal coil are in thermal contact and will reach the same final equilibrium temperature. Therefore, the total heat generated must be shared between them to raise their temperatures together.

Q: What does the area under the power-time graph represent?

The area under the power-time graph represents the total energy dissipated as heat. Since the power is constant, the area is simply a rectangle: Power multiplied by the total time.

Question

A thermocol vessel contains $0.5	ext{ kg}$ of distilled water at $30^\circ	ext{C}$. A metal coil of area $5 	imes 10^{-3}	ext{ m}^2$, number of turns $100$, mass $0.06	ext{ kg}$ and resistance $1.6\ \Omega$ is lying horizontally at the bottom of the vessel. A uniform time varying magnetic field is setup to pass vertically through the coil at time $t = 0$. The field is first increased from $0$ to $0.8	ext{ T}$ at a constant rate between $0$ and $0.2	ext{ s}$ and then decreased to zero at the same rate between $0.2$ and $0.4	ext{ s}$.
The cycle is repeated $12000$ times. Make sketches of the current through the coil and the power dissipated in the coil as a function of time for the first two cycles. Clearly indicate the magnitudes of the quantities on the axes. Assume that no heat is lost to the vessel or the surroundings. Determine the final temperature of the water under thermal equilibrium.
Specific heat of metal $= 500	ext{ J/kg-K}$ and the specific heat of water $= 4200	ext{ J/kg-K}$. Neglect the inductance of coil.

Sigma Percentile · Accepted Answer

### The Setup: A Magnetic Heater

Imagine a thermocol vessel filled with water, sitting quietly. Inside, at the bottom, lies a metal coil. Suddenly, a magnetic field starts pulsing through it, growing stronger, then weaker, over and over again. What happens? According to the laws of electromagnetism, this changing magnetic field will induce a current in the coil, turning it into a heater! Our goal is to find out exactly how much the water's temperature rises after $12,000$ of these magnetic pulses.

### Uncovering the Induced EMF

The first step is to understand the magnetic field's behavior. It increases from $0$ to $0.8	ext{ T}$ in $0.2	ext{ s}$, and then drops back to $0$ in the next $0.2	ext{ s}$. This creates a triangular wave pattern.

To find the induced EMF, we need the rate of change of this magnetic field, $\left|\frac{dB}{dt}ight|$. Since the change is linear, we can simply divide the change in the field by the time it takes:
$$ \left|\frac{dB}{dt}ight| = \frac{0.8	ext{ T}}{0.2	ext{ s}} = 4	ext{ T/s} $$

Now, we bring in Faraday's Law of Induction. The magnitude of the induced EMF ($e$) is the product of the number of turns ($N$), the area of the coil ($A$), and the rate of change of the magnetic field:
$$ e = NA\left|\frac{dB}{dt}ight| $$
$$ e = (100)(5 	imes 10^{-3}	ext{ m}^2)(4	ext{ T/s}) = 2	ext{ V} $$

### The Dance of Current and Power

With a constant EMF of $2	ext{ V}$ during each half-cycle, we can easily find the induced current using Ohm's Law. We divide the EMF by the coil's resistance ($R = 1.6\ \Omega$):
$$ i = \frac{e}{R} = \frac{2	ext{ V}}{1.6\ \Omega} = 1.25	ext{ A} $$

But wait, there's a twist! Lenz's Law tells us that the induced current will always oppose the change in magnetic flux. When the field is increasing (from $0$ to $0.2	ext{ s}$), the current flows in one direction. When the field is decreasing (from $0.2$ to $0.4	ext{ s}$), the current must reverse its direction. Thus, the current forms a square wave, alternating between $+1.25	ext{ A}$ and $-1.25	ext{ A}$.

Now, what about the power dissipated as heat? Power is given by Joule's heating formula:
$$ P = i^2 R $$
Because the current is squared, the negative sign during the second half of the cycle vanishes. The power dissipated is constant throughout the entire cycle:
$$ P = (1.25	ext{ A})^2 (1.6\ \Omega) = 2.5	ext{ W} $$

### The Grand Finale: Calorimetry

We know the coil acts as a $2.5	ext{ W}$ heater. How much total heat energy ($H$) does it produce over the $12,000$ cycles? 
One cycle takes $0.4	ext{ s}$, so the total time is:
$$ t_{total} = 12000 	imes 0.4	ext{ s} = 4800	ext{ s} $$
The total heat generated is simply power multiplied by time:
$$ H = P 	imes t_{total} = 2.5	ext{ W} 	imes 4800	ext{ s} = 12000	ext{ J} $$

Finally, we use the principle of calorimetry. This $12,000	ext{ J}$ of heat is absorbed by both the water and the metal coil, raising their temperature from $30^\circ	ext{C}$ to a final equilibrium temperature, $	heta$.
$$ H = m_w s_w (	heta - 30) + m_c s_c (	heta - 30) $$
Substituting the given values for mass and specific heat:
$$ 12000 = [0.5 	imes 4200 + 0.06 	imes 500](	heta - 30) $$
$$ 12000 = [2100 + 30](	heta - 30) $$
$$ 12000 = 2130(	heta - 30) $$
$$ 	heta - 30 = \frac{12000}{2130} \approx 5.63 $$
$$ 	heta = 35.63^\circ	ext{C} $$

The final temperature of the water is approximately **$35.6^\circ	ext{C}$**. A beautiful symphony of electromagnetism and thermodynamics!

Visualized Solution

$B$ vs $t$ Graph

$\frac{d B}{d t}$

$e = N A \frac{d B}{d t}$

$i = \frac{e}{R}$

$i$ vs $t$ Graph

$P = i^{2} R$

$P$ vs $t$ Graph

$H = P \times t_{t o t a l}$

Calorimetry Principle

Final Temperature

The Sigma Insight: Faraday's Laws of Electromagnetic Induction

The Setup

Uncovering the Induced EMF

The Dance of Current and Power

The Grand Finale

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List-I

List-II

Visualized Solution

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