Answer: (A)

Applying Faraday's law requires identifying that induced emf depends on the rate of change of magnetic flux $\Phi_B = BA$. Since the area $A = \pi r^2$, the emf is proportional to $r^2$ and the rate $\frac{dB}{dt}$. Doubling the radius increases the area by a factor of 4, and tripling the rate contributes a factor of 3, resulting in a total factor of $4 \times 3 = 12$.

Answer: (A)

Analyzing the functional dependence of motional emf $\mathcal{E} = BLv$ reveals how the variable changes combine. Substituting the new values gives $\mathcal{E}_{new} = (2B)(L/2)(4v) = 4BLv$, which is four times the original emf.

Answer: (A)

Determining the new emf requires analyzing the functional dependence of inductance $L = \frac{\mu_0 N^2 A}{\ell}$. Doubling $N$ increases $L$ by a factor of 4, while doubling $\ell$ decreases $L$ by a factor of 2. The net change is a factor of 2, so the induced emf $\mathcal{E} = L \frac{dI}{dt}$ doubles.

Answer: (A)

Evaluating the generator equation $\mathcal{E}_{max} = NBA\omega$ shows the direct proportionality to both area and angular speed. Doubling $\omega$ introduces a factor of 2, while halving $A$ introduces a factor of 0.5. The product is $2 \times 0.5 = 1$, so the emf remains unchanged.

Answer: (A)

Using Faraday's law in integral form $\oint \vec{E} \cdot d\vec{l} = -\frac{d\Phi_B}{dt}$ for a path of radius $r < R$ reveals that $E(2\pi r) = (\pi r^2) \frac{dB}{dt}$, simplifying to $E \propto r$. Halving the distance from $R/2$ to $R/4$ therefore halves the electric field magnitude.

Answer: (A)

Calculating the motional emf $\mathcal{E} = B\ell v$ requires identifying the effective length $\ell$ perpendicular to motion. Here $\ell = s$. The new emf is $B(2s)(v/2) = Bsv$, which is equal to the original emf $\mathcal{E}_0$.

Answer: (A)

Applying calculus to Faraday's law $\mathcal{E} = |\frac{d\Phi}{dt}|$ yields $\mathcal{E}(t) = 3Ct^2$. This shows the emf depends on the square of time ($\mathcal{E} \propto t^2$). Doubling the time $T$ results in a factor of $2^2 = 4$.

Answer: (A)

Analyzing the expression for a rotating rod $\mathcal{E} = \frac{1}{2}B\omega L^2$ reveals that emf is proportional to $\omega$ and $L^2$. Tripling $L$ introduces a factor of 9, and dividing $\omega$ by 3 introduces a factor of 1/3. The product is $9/3 = 3$.

Answer: (A)

Connecting emf and resistance determines the current. The emf scales with area ($\propto r^2$), while resistance scales with circumference ($\propto r$) for the same wire type. By Ohm's law, $I = \mathcal{E}/R \propto r^2/r = r$. Thus, doubling the radius doubles the current.

Answer: (A)

Applying Faraday's law shows that induced emf is proportional to the rate of change of current $\frac{dI}{dt}$. For $I(t) = kt^2$, the derivative is $2kt$, which depends linearly on time.