Log out Go to App
Go to App

Learning Materials

Features

Discover

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An LC circuit consists of a capacitor, \(C=2.50 \mu \mathrm{F},\) and an inductor, \(L=4.0 \mathrm{mH}\). The capacitor is fully charged using a battery and then connected to the inductor. An oscilloscope is used to measure the frequency of the oscillations in the circuit. Next, the circuit is opened, and a resistor, \(R\), is inserted in series with the inductor and the capacitor. The capacitor is again fully charged using the same battery and then connected to the circuit. The angular frequency of the damped oscillations in the RLC circuit is found to be \(20 \%\) less than the angular frequency of the oscillations in the LC circuit. a) Determine the resistance of the resistor. b) How long after the capacitor is reconnected in the circuit will the amplitude of the damped current through the circuit be \(50 \%\) of the initial amplitude? c) How many complete damped oscillations will have occurred in that time?

The time-varying current in an LC circuit where \(C=10.0 \mu \mathrm{F}\) is given by \(i(t)=(1.00 \mathrm{~A}) \sin (1200 . t),\) where \(t\) is in seconds. a) At what time after \(t=0\) does the current reach its maximum value? b) What is the total energy of the circuit? c) What is the inductance, \(L\) ?

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) A loop of wire \(5.00 \mathrm{~cm}\) in diameter is carrying a current of \(2.00 \mathrm{~A}\). What is the energy density of the magnetic field at its center? b) What current has to flow in a straight wire to produce the same energy density at a point \(4.00 \mathrm{~cm}\) from the wire?
See all solutions

Recommended explanations on Physics Textbooks

Linear Momentum

Read Explanation

Quantum Physics

Read Explanation

Physical Quantities And Units

Read Explanation

Torque and Rotational Motion

Read Explanation

Particle Model of Matter

Read Explanation

Force

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free