Step 1: Induced EMF in Rotating Discs

For a disc of radius $r$ rotating with angular velocity $\omega$ in a perpendicular magnetic field $B$, the induced EMF between the center and the rim is:

$$ \varepsilon = \int_0^r B (\omega x) dx = \frac{1}{2} B \omega r^2 $$

Let the EMFs of discs A and B be $\varepsilon_A = \frac{1}{2} B \omega_A r^2$ and $\varepsilon_B = \frac{1}{2} B \omega_B r^2$.

Step 2: Circuit Current and Torque

The discs are connected in a loop. The net EMF drives current $I$ through the total resistance $R$:

$$ I = \frac{\varepsilon_A – \varepsilon_B}{R} = \frac{B r^2}{2R} (\omega_A – \omega_B) $$

The magnetic torque exerted on a current-carrying disc is given by $\tau_{mag} = \int x dF = \int_0^r x (I B dx / r)$? No, considering radial current density, the integration yields:

$$ \tau_{mag} = \frac{1}{2} I B r^2 $$

For Disc B (Output), this magnetic torque overcomes the load torque $\tau$. So, $\tau = \frac{1}{2} I B r^2$.

Step 3: Calculating Efficiency

Efficiency $\eta$ is defined as Power Output divided by Power Input.

• Power Out: $P_{out} = \tau \omega_B$
• Power In: Power supplied to Disc A. The motor must overcome the magnetic counter-torque on A. Since the same current $I$ flows through A, the torque on A is also $\tau_A = \frac{1}{2} I B r^2 = \tau$. Thus, $P_{in} = \tau \omega_A$.

$$ \eta = \frac{P_{out}}{P_{in}} = \frac{\tau \omega_B}{\tau \omega_A} = \frac{\omega_B}{\omega_A} $$

Now we express $\omega_A$ in terms of $\omega_B$ and constants. From the torque equation:

$$ \tau = \frac{1}{2} B r^2 \left( \frac{B r^2 (\omega_A – \omega_B)}{2R} \right) = \frac{B^2 r^4}{4R} (\omega_A – \omega_B) $$ $$ \implies \omega_A – \omega_B = \frac{4 R \tau}{B^2 r^4} $$ $$ \implies \omega_A = \omega_B + \frac{4 R \tau}{B^2 r^4} $$

Substituting this back into the efficiency formula: