Solution to Question 4
1. Kinematic Analysis
The bobbin rolls without slipping on the horizontal table. Let $v_0$ be the velocity of the center of mass ($O$) and $\omega$ be the angular velocity of the bobbin.
- Instantaneous Axis of Rotation (IAR): Since there is no slipping, the point of contact with the table is the IAR (velocity = 0).
- Velocity of Wire at Q: The wire at $Q$ is pulled with velocity $u$. The wire unwinds from the bottom of the inner cylinder.
- Relating $u$ and $\omega$: The velocity of the bottom of the inner cylinder relative to the IAR is $u$. The distance from the IAR (ground) to this point is $(R – r)$.
$$ u = \omega (R – r) \implies \omega = \frac{u}{R – r} $$
(a) Moduli of Induced Electromotive Forces (EMF)
The problem asks for the EMF induced in the coils due to the unwrapping action (rotational flux change). The rate at which area is swept by the unwinding wire is $\frac{dA}{dt} = \frac{1}{2} \omega (\text{radius})^2$.
Outer Coil (Radius $R$)
$$ \mathscr{E}_{\text{Outer}} = \frac{1}{2} B \omega R^2 = \frac{B u R^2}{2(R – r)} $$
Inner Coil (Radius $r$)
$$ \mathscr{E}_{\text{Inner}} = \frac{1}{2} B \omega r^2 = \frac{B u r^2}{2(R – r)} $$
(b) Voltmeter Reading
The voltmeter measures the potential difference between point $P$ (ground) and the sliding contact $S$ (wire exit). Using the concept of **Motional EMF relative to the IAR**:
$$ V = \frac{1}{2} B \omega (d_{\text{end}}^2 – d_{\text{start}}^2) $$
- Start Point ($P$): Distance from IAR $d_{\text{start}} = 0$.
- End Point ($S$): Distance from IAR $d_{\text{end}} = R – r$.
$$ V = \frac{1}{2} B \omega (R – r)^2 $$ Substituting $\omega$: $$ V = \frac{1}{2} B \left( \frac{u}{R – r} \right) (R – r)^2 = \frac{B(R – r)u}{2} $$
Deep Dive: Why Subtraction $(\mathscr{E}_{\text{Outer}} – \mathscr{E}_{\text{Inner}})$ Fails
A very common logical trap is to think: “The total effect is the combination of two coils, so shouldn’t the voltage be the difference of their individual EMFs?”
The answer is No, because the bobbin is rolling (Translation + Rotation), not just rotating about a fixed center.
1. The Missing Piece: Translational Velocity
The formula $\mathscr{E} = \frac{1}{2} B \omega R^2$ used in part (a) assumes the center of rotation is stationary. However, the voltmeter is stationary on the ground, measuring a bobbin that is speeding sideways. We must account for the translational velocity.
2. Visualizing the Velocity Profile
The Induced EMF is determined by the “area” under the velocity graph along the wire ($V = \int B v \, dl$).
- Correct Method (IAR): Measuring from the ground ($P$) to the inner coil bottom ($S$), the velocity rises from $0$ to $u$. This forms a triangle in the velocity-distance graph.
$\text{Area} \propto \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} B \omega (R-r)^2$. - Subtraction Method (Trapezoid Error): If you try to subtract the inner coil’s EMF, you are assuming its contribution is small (like a small triangle starting from zero). In reality, the inner coil is “riding” on the moving bobbin. Its velocity is very high (ranging from $\omega R$ down to $u$). This region is a trapezoid, not a small triangle.
3. Mathematical Proof
We can prove this by summing the potentials step-by-step from the ground up:
Step A: Ground to Center ($V_O – V_P$)
Velocity goes from $0$ to $\omega R$.
$$ V_O – V_P = \frac{1}{2} B \omega R^2 $$
Step B: Center to Inner Exit ($V_O – V_S$)
Here we integrate the high velocity region.
$$ V_O – V_S = \int_{R-r}^{R} B (\omega y) dy = \frac{1}{2} B \omega [R^2 – (R-r)^2] $$
Step C: Total Reading ($V_S – V_P$)
$$ V_S – V_P = (V_O – V_P) – (V_O – V_S) $$
$$ = \frac{1}{2} B \omega R^2 – \frac{1}{2} B \omega [R^2 – (R-r)^2] $$
$$ = \frac{1}{2} B \omega (R-r)^2 $$
This confirms that simple subtraction fails because it ignores the large potential drop caused by the translational motion of the inner coil.