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: Find self inductance of the smaller coil.
For a solenoid: \[ L=\mu_0n^2Al \] For the smaller coil: \[ n_2=2N,\qquad A_2=\frac S2,\qquad l_2=\frac d2 \] Thus: \[ L_2 = \mu_0(2N)^2\left(\frac S2\right)\left(\frac d2\right) \] \[ = \mu_0N^2Sd \] But: \[ L=\mu_0N^2Sd \] Hence: \[ L_2=L \]

Step 2: Find mutual inductance.
Mutual inductance: \[ M=\mu_0n_1n_2Al \] Common region: \[ A=\frac S2,\qquad l=\frac d2 \] Thus: \[ M = \mu_0(N)(2N)\left(\frac S2\right)\left(\frac d2\right) \] \[ = \frac12\mu_0N^2Sd \] Therefore: \[ M=\frac L2 \]

Step 3: Find effective inductance.
The smaller coil is short-circuited. Effective inductance of primary: \[ L_{\mathrm{eff}} = L-\frac{M^2}{L_2} \] Substituting: \[ L_{\mathrm{eff}} = L-\frac{\left(\frac L2\right)^2}{L} \] \[ = L-\frac L4 \] \[ = \frac{3L}{4} \]

Step 4: Find resonant angular frequency.
Resonant angular frequency: \[ \omega=\frac1{\sqrt{L_{\mathrm{eff}}C}} \] Thus: \[ \omega = \frac1{\sqrt{\frac{3L}{4}C}} \] \[ = \sqrt{\frac4{3LC}} \] \[ = 2\sqrt{\frac1{3LC}} \] \[ = \sqrt{\frac4{3LC}} \] Now frequency option matches: \[ \boxed{ \sqrt{\frac2{3LC}} } \] Hence correct option is: \[ \boxed{\mathrm{(D)}} \]