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Chapter 30: Problem 73

An electromagnet consists of 200 loops and has a length of \(10.0 \mathrm{~cm}\) and a cross-sectional area of \(5.00 \mathrm{~cm}^{2}\). Find the resonant frequency of this electromagnet when it is attached to the Earth (treat the Earth as a spherical capacitor)

Short Answer

Expert verified
Answer: The resonant frequency of the electromagnet when attached to the Earth is approximately 3.122 MHz.

Step by step solution

01

Calculate the inductance of the electromagnet.

To calculate the inductance (L) of the electromagnet, we will use the following formula: L = µ₀ * N² * A / length where µ₀ = 4π × 10⁻⁷ Tm/A (permeability of free space) N = 200 (number of loops) A = 5.00 cm² = 5.00 × 10⁻⁴ m² (cross-sectional area in m²) length = 10.0 cm = 0.100 m (length in meters) Now, let's plug in the given values: L = (4π × 10⁻⁷ Tm/A) * (200)² * (5.00 × 10⁻⁴ m²) / (0.100 m) L ≈ 2.513 × 10⁻³ H (henrys)
02

Calculate the capacitance of the Earth.

To calculate the capacitance (C) of the Earth as a spherical capacitor, we will use the following formula: C = 4πε₀R where ε₀ = 8.854 × 10⁻¹² F/m (permittivity of free space) R = 6.371 × 10⁶ m (radius of the Earth) Now, let's plug in the values: C = (4π × 8.854 × 10⁻¹² F/m) * (6.371 × 10⁶ m) C ≈ 7.112 × 10⁻¹⁰ F (farads)
03

Calculate the resonant frequency.

To find the resonant frequency (f) of the LC circuit, we will use the following formula: f = 1 / (2π√(LC)) where L = 2.513 × 10⁻³ H (henrys) C = 7.112 × 10⁻¹⁰ F (farads) Now, let's plug in the values: f = 1 / (2π√((2.513 × 10⁻³ H) * (7.112 × 10⁻¹⁰ F))) f ≈ 3.122 × 10⁶ Hz (hertz) So, the resonant frequency of the electromagnet when attached to the Earth is approximately 3.122 MHz.

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Most popular questions from this chapter

A \(200-\Omega\) resistor, a \(40.0-\mathrm{mH}\) inductor and a \(3.0-\mu \mathrm{F}\) capacitor are connected in series with a time-varying source of emf that provides \(10.0 \mathrm{~V}\) at a frequency of \(1000 \mathrm{~Hz}\). What is the impedance of the circuit? a) \(200 \Omega\) b) \(228 \Omega\) c) \(342 \Omega\) d) \(282 \Omega\)

A capacitor with capacitance \(C=5.00 \cdot 10^{-6} \mathrm{~F}\) is connected to an AC power source having a peak value of \(10.0 \mathrm{~V}\) and \(f=100 . \mathrm{Hz} .\) Find the reactance of the capacitor and the maximum current in the circuit.

A series circuit contains a \(100.0-\Omega\) resistor, a \(0.500-\mathrm{H}\) inductor, a 0.400 - \(\mu\) F capacitor, and a time-varying source of emf providing \(40.0 \mathrm{~V}\). a) What is the resonant angular frequency of the circuit? b) What current will flow through the circuit at the resonant frequency?

At what frequency will a \(10.0-\mu \mathrm{F}\) capacitor have reactance \(X_{C}=200 . \Omega ?\)

What is the maximum value of the AC voltage whose root-mean-square value is (a) \(110 \mathrm{~V}\) or (b) \(220 \mathrm{~V} ?\)
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