Area of loop, \( A = 1.0 \text{ m}^2 \)
Magnetic field: \( B = \sin(100t) \)
Magnetic flux through loop:
\[ \Phi = BA = \sin(100t) \times 1 \]
\[ \Phi = \sin(100t) \]
Induced emf (Faraday’s law):
\[ e = -\frac{d\Phi}{dt} \]
\[ e = -100\cos(100t) \]
Magnitude:
\[ e = 100\cos(100t) \]
Instantaneous current:
\[ i = \frac{e}{R} = \frac{100\cos(100t)}{100} \]
\[ i = \cos(100t) \]
Instantaneous power dissipated:
\[ P = i^2R \]
\[ P = \cos^2(100t) \times 100 \]
Average value of \( \cos^2 \) over one period is \( \frac{1}{2} \).
\[ P_{avg} = 100 \times \frac{1}{2} = 50 \]
Angular frequency \( \omega = 100 \)
\[ T = \frac{2\pi}{\omega} = \frac{2\pi}{100} \]
Energy dissipated in one period:
\[ E = P_{avg} \times T \]
\[ E = 50 \times \frac{2\pi}{100} \]
\[ E = \pi \text{ J} \]
Therefore correct answer is (C).
Electromagnetic induction is one of the most important and conceptually rich topics in Physics for JEE Main, JEE Advanced and NEET. It connects magnetism with electricity and forms the foundation of electric generators, transformers and alternating current theory.
The core principle used in this question is Faraday’s Law of Electromagnetic Induction. It states that whenever the magnetic flux linked with a circuit changes with time, an emf is induced in the circuit. Mathematically,
\[ e = -\frac{d\Phi}{dt} \]
The negative sign represents Lenz’s law which states that the induced current always opposes the cause producing it. This opposition ensures conservation of energy.
Magnetic flux is defined as:
\[ \Phi = BA\cos\theta \]
Where B is magnetic field, A is area and θ is angle between field and area vector. In this question the loop is perpendicular to magnetic field, so θ = 0 and cosθ = 1.
When magnetic field varies sinusoidally, induced emf also becomes sinusoidal. This directly connects electromagnetic induction with alternating current (AC) theory.
If:
\[ B = B_0 \sin(\omega t) \]
Then:
\[ e = \omega B_0 A \cos(\omega t) \]
This shows induced emf amplitude is proportional to angular frequency. Higher frequency means greater induced emf.
In resistive loop, induced current is:
\[ i = \frac{e}{R} \]
Instantaneous power dissipated:
\[ P = i^2R \]
For sinusoidal current:
\[ i = I_0 \cos(\omega t) \]
\[ P = I_0^2 R \cos^2(\omega t) \]
Average of cos² over one complete cycle:
\[ \langle \cos^2 \rangle = \frac{1}{2} \]
This result is extremely important in AC circuit theory and appears frequently in JEE Main.
Thus average power:
\[ P_{avg} = \frac{I_0^2 R}{2} \]
Thermal energy dissipated in one cycle:
\[ E = P_{avg}T \]
Where:
\[ T = \frac{2\pi}{\omega} \]
Students must clearly understand difference between instantaneous power and average power. Many students directly substitute peak values incorrectly.
Core Concepts Involved:
• Faraday’s Law
• Lenz’s Law
• Magnetic flux
• Sinusoidal variation
• Average value of trigonometric squares
• Resistive power dissipation
Important Formulas for Exams:
\[ e = -\frac{d\Phi}{dt} \]
\[ \Phi = BA\cos\theta \]
\[ I_{rms} = \frac{I_0}{\sqrt{2}} \]
\[ P_{avg} = I_{rms}^2 R \]
\[ T = \frac{2\pi}{\omega} \]
Shortcut Trick:
If emf amplitude is \(E_0\), then current amplitude \(I_0 = \frac{E_0}{R}\). Average power becomes:
\[ P_{avg} = \frac{E_0^2}{2R} \]
This shortcut directly gives answer without writing time dependent expressions.
Exam Relevance:
Electromagnetic induction contributes 2–3 questions in JEE Main almost every year. In JEE Advanced, it is frequently combined with rotation, mechanics or electrostatics.
Common variations include:
• Rotating rod in magnetic field
• Sliding conductor on rails
• Induced charge calculation
• Induced emf in moving loop
• Power dissipation problems
Common Mistakes Students Make:
• Forgetting minus sign conceptually
• Using wrong period formula
• Forgetting average of sin² or cos² is 1/2
• Confusing rms value with peak value
• Not squaring current while calculating power
Conceptual Depth:
This problem demonstrates direct conversion of magnetic energy variation into electrical energy which ultimately converts into heat. This energy conversion chain is fundamental in generators.
AC heating effect is used in electric heaters, induction cooktops and transformers. In induction heating, rapidly varying magnetic field induces strong currents in metal producing heating effect.
Mastering this topic builds strong foundation for AC circuits, transformers and electromagnetic waves.
For JEE Main and Advanced, understanding derivations and average value concepts is extremely important. Practice problems involving rms values and power calculations to gain speed and accuracy.