Question

Consider a circuit consisting of a capacitor of capacitance \(C\) and a coil with \(N\) turns per unit length, cross sectional area \(S\) and length \(d\), where \[ d^2\gg S \] There is another coil of length \[ \frac d2 \] cross sectional area \[ \frac S2 \] and \(2N\) turns per unit length completely inside the larger coil, as shown in the figure. The ends of this smaller coil are connected with each other by an insulated conducting wire. The self-inductance of the larger coil is \(L\). Neglecting edge effects and all the Ohmic resistances, the resonant frequency of the circuit is:

• A. \(\dfrac4{\sqrt{15LC}}\)
• B. \(\dfrac6{\sqrt{5LC}}\)
• C. \(\dfrac2{\sqrt{3LC}}\)
• D. \(\sqrt{\dfrac2{3LC}}\)

Hint:

For a short-circuited secondary coil: \[ L_{\mathrm{eff}}=L_1-\frac{M^2}{L_2} \]

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
When an inner coil is shorted, it acts as a secondary circuit that reduces the effective inductance of the primary coil due to mutual inductance. The resonant frequency is \( 1/\sqrt{L_{eff}C} \).
Step 3: Detailed Explanation:
1) Self and Mutual Inductances:
Primary \( L = \mu_0 N^2 S d \).
Secondary \( L_2 = \mu_0 (2N)^2 (S/2) (d/2) = \mu_0 (4N^2) (S/4) d = \mu_0 N^2 S d = L \).
Mutual \( M = \mu_0 N (2N) (S/2) (d/2) = \mu_0 N^2 S d / 2 = L/2 \).
2) Effective Inductance:
Equation for shorted secondary: \( L_2 \frac{di_2}{dt} + M \frac{di_1}{dt} = 0 \implies \frac{di_2}{dt} = -\frac{M}{L_2} \frac{di_1}{dt} \).
Voltage in primary: \( V = L \frac{di_1}{dt} + M \frac{di_2}{dt} = (L - \frac{M^2}{L_2}) \frac{di_1}{dt} \).
\( L_{eff} = L - \frac{(L/2)^2}{L} = L - L/4 = 3L/4 \).
3) Resonant frequency:
\( \omega = \frac{1}{\sqrt{L_{eff}C}} = \frac{1}{\sqrt{(3L/4)C}} = \frac{2}{\sqrt{3LC}} \).
Step 4: Final Answer:
Matches (C).