44. Thermal Power of Entering Loop
Step 1: Motional EMF
As the sinusoidal loop enters the magnetic field $B$ with velocity $v$, the change in flux creates a motional EMF. The vertical span of the wire at the boundary determines the instantaneous EMF.
The sinusoid is $y(t) = a_0 \sin(\omega t)$ where $\omega = \frac{2\pi}{T}$.
The induced EMF is:
$$ \varepsilon = B v y(t) = B v a_0 \sin(\omega t) $$Step 2: Circuit Current (LR Circuit)
The loop has resistance $R$ and inductance $L$. The current satisfies:
$$ L \frac{di}{dt} + iR = \varepsilon_{peak} \sin(\omega t) $$This is an AC circuit problem. The impedance is $Z = \sqrt{R^2 + (\omega L)^2}$.
The amplitude of the current is:
$$ I_0 = \frac{\varepsilon_{peak}}{Z} = \frac{B v a_0}{\sqrt{R^2 + \omega^2 L^2}} $$Step 3: Average Power
Average thermal power dissipated in the resistor is $P = \frac{1}{2} I_0^2 R$ (using RMS current $I_0/\sqrt{2}$).
$$ P_{avg} = \frac{1}{2} \left( \frac{B v a_0}{\sqrt{R^2 + \omega^2 L^2}} \right)^2 R $$ $$ P_{avg} = \frac{1}{2} \frac{B^2 v^2 a_0^2 R}{R^2 + \frac{4\pi^2 L^2}{T^2}} $$Step 4: Final Expression
Multiply numerator and denominator by $T^2$ to clear the fraction:
$$ P_{avg} = \frac{B^2 v^2 a_0^2 R T^2}{2 (T^2 R^2 + 4\pi^2 L^2)} $$
$$ P_{avg} = \frac{(B v a_0)^2 R T^2}{2(T^2 R^2 + 4\pi^2 L^2)} $$