Question

A metallic disc of radius 'a' mounted on a rigid axle is rotating with a constant angular velocity $\omega$ inside a long solenoid of inductance L whose two ends are connected to the rotating disc by wo bruch constants as shown. The total resistance of the whole circuit is R. Then

• A. The induced emf between terminals P and Q is $\frac{\mu_0 Ni\omega a^2}{2l}$, when current in the coil is i.
• B. Current in the coil can increase or decrease with time depending upon the value of $\omega$.
• C. $\omega_{min}$ for increasing the current in the coil always is $\frac{lR}{2\mu_0 Na^2}$
• D. $\omega_{min}$ for increasing the current in the coil always is $\frac{2lR}{2\mu_0 Na^2}$

Answer 1

Answer: (A), (B)

Options (A) and (B)

Answer 2

1. Induced emf in the disc:
   A point at a distance r from the center has speed ωr so a small emf dε = Bωr dr is induced. Integrate from r = 0 to a:
   ε = ∫₀ᵃ Bωr dr = ½ Bωa².

2. Magnetic field from the solenoid:
   Inside a long solenoid, B = μ₀(N/l) i. Hence,
   ε = ½ (μ₀(N/l)i) ωa² = (μ₀N ω a²)/(2l) i.
   This exactly matches option (A).

3. Circuit equation:
   The loop (with inductance L and resistance R) follows
   L(d i/dt) + Ri = (μ₀N ω a²)/(2l) i.
   Thus,
   d i/dt = i [(μ₀N ω a²)/(2l) – R] / L.
   So if (μ₀N ω a²)/(2l) > R the current grows, whereas if it is less than R the current decays.
   This is what option (B) states.

4. Threshold angular velocity:
   Setting the condition for self-excitation,
   (μ₀N ω a²)/(2l) = R ⟹ ω = (2lR)/(μ₀N a²).
   Neither option (C) nor (D) matches this expression (option (C) gives ω = lR/(2μ₀Na²), and (D) simplifies to ω = lR/(μ₀Na²)). Hence options (C) and (D) are incorrect.