42. Power Dissipated in Induced Loop
Step 1: Analyze First Loop
The loop behaves as an LR circuit driven by an induced EMF $\varepsilon$. The phase shift $\phi$ between flux (source current) and induced current is given by $\tan \phi = \frac{\omega L}{R}$.
Given $\phi = 45^\circ$, so $\tan 45^\circ = 1 \implies \omega L = R$.
The power dissipated is:
$$ p = \frac{\varepsilon_{rms}^2}{Z^2} R = \frac{\varepsilon_{rms}^2}{R^2 + (\omega L)^2} R $$Since $\omega L = R$, $Z^2 = 2R^2$.
$$ p = \frac{\varepsilon_{rms}^2}{2R^2} R = \frac{\varepsilon_{rms}^2}{2R} \implies \varepsilon_{rms}^2 = 2pR $$Step 2: Analyze Second Loop
The new loop has identical shape and size, so its inductance $L$ and the induced EMF $\varepsilon$ remain the same. The resistivity is $\eta$ times higher, so the new resistance is $R’ = \eta R$.
The new power $p’$ is:
$$ p’ = \frac{\varepsilon_{rms}^2}{R’^2 + (\omega L)^2} R’ $$Step 3: Calculation
Substitute $R’ = \eta R$ and $\omega L = R$:
$$ p’ = \frac{\varepsilon_{rms}^2}{(\eta R)^2 + R^2} (\eta R) = \varepsilon_{rms}^2 \frac{\eta R}{R^2(\eta^2 + 1)} = \frac{\varepsilon_{rms}^2}{R} \frac{\eta}{\eta^2 + 1} $$Substitute $\frac{\varepsilon_{rms}^2}{R} = 2p$ from Step 1:
$$ p’ = (2p) \frac{\eta}{\eta^2 + 1} $$
$$ P_{new} = \frac{2\eta p}{\eta^2 + 1} $$