Q28
jee_main_2026_21_jan_morning
Faraday's Law
A conducting circular loop of area 1.0 \, mathrmm^2 is placed perpendicular to a magnetic field which varies as B = sin(100 \, t)text Tesla. If the resistance of the loop is 100 \, Omega, then the average thermal energy dissipated in the loop in one period is ____ J.
Solution
### Related Formula
phi = B cdot A
E = -fracdphidt
P = fracE^2R
### Core Logic
Given area of the loop, A = 1text m^2 and magnetic field B = sin(100t).
The magnetic flux passing through the loop is:
phi = B cdot A = sin(100t) times 1 = sin(100t)
Induced EMF E = left| fracdphidt right| = 100cos(100t)
### Step 1: Calculating Power and Energy
Instantaneous power P = fracE^2R = frac100^2 cos^2(100t)100 = 100cos^2(100t).
Thermal energy dissipated in one time period T:
Q = int_0^T P \, dt = int_0^T 100cos^2(100t) \, dt
The angular frequency omega = 100text rad/s, so time period T = frac2piomega = frac2pi100 = fracpi50text sec.
Q = 100 int_0^pi/50 cos^2(100t) \, dt = 100 int_0^pi/50 frac1 + cos(200t)2 \, dt
Q = 50 left[ t + fracsin(200t)200 right]_0^pi/50
Q = 50 left( fracpi50 - 0 right) = pitext Joules
### Pattern Recognition
For a sinusoidal signal, the integral of cos^2(omega t) over one full period T is always T/2. Thus, int P \, dt = P_textmax times fracT2.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Induction
Class 12 Physics: Alternating Current
