JEE MainPhysicsAlternating Current
The search coil of a metal detector has an inductance L and is connected to a capacitor C , forming an oscillator that resonates at a frequency f₀ . When a metallic object is brought near the coil, eddy currents induced in the metal cause the effective inductance of the search coil to decrease to 0.81 L . What is the new resonant frequency of the circuit in terms of f₀ ?
Options
- A0.9 f₀
- B0.81 f₀
- C0.9 f₀
- D10 9 f₀
Correct answer
D. 10 9 f₀
Step-by-step solution
The initial resonant frequency of the LC circuit is given by: f₀ = 1 2 LC When the metallic object is brought near the coil, the new effective inductance becomes L' = 0.81 L . The new resonant frequency is: f' = 1 2 L'C = 1 2 0.81 LC Since 0.81 = 0.9 , we have: f' = 1 0.9 ( 1 2 LC ) = 10 9 f₀ Answer: 10 9 f₀