The current flowing in a solenoid that is \(20.0 \mathrm{~cm}\) long and has a radius of \(2.00 \mathrm{~cm}\) and 500 turns decreases from \(3.00 \mathrm{~A}\) to \(1.00 \mathrm{~A}\) in \(0.1 .00 \mathrm{~s} .\) Determine the magnitude of the induced electric field inside the solenoid \(1.00 \mathrm{~cm}\) from its center.

Short Answer

Expert verified
Question: Calculate the induced electric field inside a solenoid at a distance of 1.00 cm from its center when the current decreases from 3.00 A to 1.00 A in 0.100 s. The solenoid has a total of 500 turns and is 20.0 cm long with a radius of 2.00 cm. Answer: To find the induced electric field, follow these steps: 1. Calculate the change in magnetic field inside the solenoid using the formula \(\Delta B = \mu_0 * \frac{N}{L} * (I_{initial} - I_{final})\) 2. Calculate the change in magnetic flux using the formula \(\Delta \Phi = A * \Delta B\), with \(A = \pi * r^2\) 3. Calculate the induced EMF using Faraday's Law, \(|EMF| = \left|\frac{\Delta \Phi}{\Delta t}\right|\) 4. Calculate the induced electric field using the formula \(E = \frac{EMF}{2 \pi r}\) After calculating each step, you'll find the magnitude of the induced electric field inside the solenoid at a distance of 1.00 cm from its center.

Step by step solution

01

Calculate the change in magnetic field inside the solenoid

To find the change in magnetic field, we first need to know the magnetic field inside the solenoid when the current is changing. The formula for the magnetic field inside a solenoid is given by: \(B = \mu_0 * \frac{N}{L} * I\) Where \(B\) is the magnetic field, \(\mu_0\) is the permeability of free space (\(4\pi \times 10^{-7} Tm/A\)), \(N\) is the number of turns, \(L\) is the length of the solenoid, and \(I\) is the current flowing through the solenoid. Here, we are given that the current decreases from \(3.00 A\) to \(1.00 A\). Therefore, the change in the magnetic field will be: \(\Delta B = B_{initial} - B_{final} = \mu_0 * \frac{N}{L} * (I_{initial} - I_{final})\)
02

Calculate the change in magnetic flux

Now, we can calculate the change in magnetic flux using the change in the magnetic field. The formula to find the magnetic flux is given by: \(\Delta \Phi = A * \Delta B\) Where \(\Delta \Phi\) is the change in magnetic flux, \(A\) is the cross-sectional area of the solenoid, and \(\Delta B\) is the change in magnetic field. For a solenoid with a radius of \(r\), the cross-sectional area is given by: \(A = \pi r^2\) In this case, the radius is not given directly but we can calculate it by considering that we need to find the induced electric field \(1.00 cm\) from the center of the solenoid. Since the total radius of the solenoid is \(2.00 cm\), the effective radius for our calculation will be \(2.00 cm - 1.00 cm = 1.00 cm\). So, \(A = \pi * (0.01 m)^2\) Now, we can find the change in magnetic flux using the above formula.
03

Calculate the induced EMF using Faraday's Law

Faraday's Law of electromagnetic induction states that the induced EMF (Electromotive Force) is equal to the negative rate of change of magnetic flux. Mathematically, it is given by: \(EMF = -\frac{d\Phi}{dt}\) However, in this problem, we are only interested in the magnitude of the induced EMF. So, we can write: \(|EMF| = \left|\frac{\Delta \Phi}{\Delta t}\right|\) Where \(\Delta t\) is the time taken for the change in current, which is given to be \(0.100 s\) in this problem. Now, we can find the induced EMF using the change in magnetic flux calculated in Step 2.
04

Calculate the induced electric field

Finally, we can calculate the induced electric field using the induced EMF. The formula to find the electric field inside a solenoid is given by: \(E = \frac{EMF}{2 \pi r}\) Where \(E\) is the induced electric field, \(EMF\) is the induced electromotive force, and \(r\) is the distance from the center of the solenoid. In this case, the distance from the center of the solenoid is given to be \(1.00 cm\). So, we can plug in the values and find the induced electric field.

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.

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

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Faraday's Law of electromagnetic induction
Faraday's Law of electromagnetic induction is a fundamental principle that explains how electric currents can be produced by changing magnetic fields. This law is the basic operation behind many electrical generators and transformers. According to Faraday's Law, an electromotive force (EMF) is induced in a circuit whenever there is a change in magnetic flux through that circuit. The size of the induced EMF is directly proportional to the rate of change of the magnetic flux.

Mathematically, Faraday's Law is expressed as:
\[EMF = -N \frac{d\Phi}{dt}\]
Here, \(EMF\) represents the induced electromotive force in volts, \(N\) is the number of turns in the solenoid or coil through which the flux is changing, \(d\Phi\) is the change in magnetic flux in webers over time \(dt\) in seconds. The negative sign indicates the direction of the induced EMF (and the resulting current, if the circuit is closed) which is given by Lenz's Law. This law states that the induced current will flow in a direction that opposes the change in magnetic flux that produced it, which is also a consequence of the law of conservation of energy.

In the context of the exercise, Faraday's Law is used to calculate the magnitude of the induced EMF when the current in the solenoid changes over time.
Magnetic flux
The concept of magnetic flux (\(\Phi\)) is central to understanding Faraday's Law of electromagnetic induction. Magnetic flux is a measure of the total magnetic field (B) passing through a certain area (A). It's comparable to the amount of water flowing through a pipe - the more water, the greater the flow. Similarly, the greater the magnetic field that intersects a surface area, the larger the magnetic flux.

Mathematically, magnetic flux is given by the equation:
\[\Phi = B \cdot A \cdot \cos(\theta)\]
Where \(\Phi\) is the magnetic flux in webers (Wb), \(B\) is the magnetic field in teslas (T), \(A\) is the area in square meters (m^2) through which the field lines pass, and \(\theta\) is the angle between the magnetic field lines and the normal (perpendicular) to the area. For a solenoid with a uniform magnetic field and an area perpendicular to the field lines, the cosine term is equal to one, simplifying the calculation of magnetic flux.

In the given problem, the magnetic flux change is key to determining the EMF, and subsequently the induced electric field inside the solenoid, as the current changes over time.
Electromotive force (EMF)
The electromotive force, or EMF, is not really a force per se. Despite its name, EMF is a measure of energy provided per unit charge, measured in volts (V). It is the work done on a charge when it completes a circuit, a potential difference that can push electric charges through a conductor. In the context of electromagnetic induction, EMF is the voltage generated when the magnetic environment around a conductor or coil is changing.

When discussing induced EMF as a result of a change in magnetic flux, it's essential to understand that this induced voltage is what drives the flow of current in a closed circuit. The magnitude of the induced EMF can be influenced by several factors, such as the rate of change in magnetic flux, the number of turns in the coil, and the cross-sectional area of the coil.

For the problem provided, using Faraday's Law, we find the magnitude of the induced EMF once we have the change in magnetic flux. The calculation disregards the direction of the induced EMF, focusing only on its absolute value to determine the strength of the induced electric field.
Magnetic field inside a solenoid
A solenoid is a coil of wire that produces a uniform magnetic field in its interior when electric current passes through it. This magnetic field is strongest in the center and aligns along the length of the coil. The strength of the magnetic field inside a solenoid is determined by several factors: the current flowing through the coils (I), the number of turns of the wire (N), the length of the solenoid (L), and the permeability of free space (\(\mu_0\)).

The formula to calculate the magnetic field inside an ideal solenoid is:
\[B = \mu_0 * \frac{N}{L} * I\]
In this expression, \(\mu_0\) is known as the magnetic constant, with a value of about \(4\pi \times 10^{-7} Tm/A\). By using the initial and final currents, we can determine the change in the magnetic field (\(\Delta B\)), which, when multiplied by the cross-sectional area of the solenoid, gives us the change in magnetic flux (\(\Delta \Phi\)). Knowing this, we can apply Faraday's Law to find the EMF and then the induced electric field at a specific location inside the solenoid.

In the given problem, the magnetic field calculation is pivotal in determining the electric field induced due to the change in the current. As the current flowing through the solenoid's coils changes, it alters the magnetic field inside, leading to electromagnetic induction.

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

A house with a south-facing roof has photovoltaic panels on the roof. The photovoltaic panels have an efficiency of \(10.0 \%\) and occupy an area with dimensions \(3.00 \mathrm{~m}\) by \(8.00 \mathrm{~m} .\) The average solar radiation incident on the panels is \(300 . \mathrm{W} / \mathrm{m}^{2}\), averaged over all conditions for a year. How many kilowatt hours of electricity will the solar panels generate in a 30 -day month?

The most intense beam of light that can propagate through dry air must have an electric field whose maximum amplitude is no greater than the breakdown value for air: \(E_{\max }^{\operatorname{air}}=3.0 \cdot 10^{6} \mathrm{~V} / \mathrm{m},\) assuming that this value is unaffected by the frequency of the wave. a) Calculate the maximum amplitude the magnetic field of this wave can have. b) Calculate the intensity of this wave. c) What happens to a wave more intense than this?

At the surface of the Earth, the Sun delivers an estimated \(1.00 \mathrm{~kW} / \mathrm{m}^{2}\) of energy. Suppose sunlight hits a \(10.0 \mathrm{~m}\) by \(30.0 \mathrm{~m}\) roof at an angle of \(90.0^{\circ}\) a) Estimate the total power incident on the roof. b) Find the radiation pressure on the roof.

Alice made a telephone call from her home telephone in New York to her fiancé stationed in Baghdad, about \(10,000 \mathrm{~km}\) away, and the signal was carried on a telephone cable. The following day, Alice called her fiancé again from work using her cell phone, and the signal was transmitted via a satellite \(36,000 \mathrm{~km}\) above the Earth's surface, halfway between New York and Baghdad. Estimate the time taken for the signals sent by (a) the telephone cable and (b) via the satellite to reach Baghdad, assuming that the signal speed in both cases is the same as speed of light, \(c .\) Would there be a noticeable delay in either case?

If two communication signals were sent at the same time to the Moon, one via radio waves and one via visible light, which one would arrive at the Moon first?

See all solutions

Recommended explanations on Physics Textbooks

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