When a magnet in an MRI is abruptly shut down, the magnet is said to be quenched. Quenching can occur in as little as \(20.0 \mathrm{~s}\). Suppose a magnet with an initial field of \(1.20 \mathrm{~T}\) is quenched in \(20.0 \mathrm{~s},\) and the final field is approximately zero. Under these conditions, what is the average induced potential difference around a conducting loop of radius \(1.00 \mathrm{~cm}\) (about the size of a wedding ring) oriented perpendicular to the field?

Short Answer

Expert verified
Answer: The average induced potential difference around the conducting loop is approximately \(1.89 \times 10^{-5}\) volts.

Step by step solution

01

Define the given variables and constants

We are given: Initial magnetic field, \(B_1 = 1.20\) T Final magnetic field, \(B_2 \approx 0\) T Time to quench the magnet, \(t = 20.0\) s Radius of the conducting loop, \(r = 1.00\) cm \(= 0.01\) m
02

Define the formula for Faraday's Law

Faraday's Law states that the induced electromotive force (EMF) is equal to the rate of change of magnetic flux through a closed loop. $$ \varepsilon = - \frac{\Delta \Phi}{\Delta t} $$
03

Calculate the change in magnetic flux

The magnetic flux \(\Phi\) through a loop of area A perpendicular to the magnetic field is given by: $$ \Phi = B\cdot A $$ Since the loop is a circle, its area can be calculated using $$ A = \pi r^2 $$ The change in magnetic flux \(\Delta \Phi\) is given by the difference between the initial and final fluxes: $$ \Delta \Phi = \Phi_2 - \Phi_1 = B_2A - B_1A $$ As \(B_2 \approx 0\), $$ \Delta \Phi = -B_1A = -1.20 T \cdot \pi (0.01 m)^2 $$
04

Calculate the average induced potential difference

Now, we can calculate the average potential difference (induced EMF) using Faraday's Law: $$ \varepsilon = - \frac{\Delta \Phi}{\Delta t} = \frac{[1.20 \mathrm{T} \cdot \pi (0.01 \mathrm{m})^2]}{20.0 \mathrm{s}} $$ Calculating the value, $$ \varepsilon \approx 1.89 \times 10^{-5} V $$ So, the average induced potential difference around the conducting loop is approximately \(1.89 \times 10^{-5}\) volts.

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!

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 motor has a single loop inside a magnetic field of magnitude \(0.87 \mathrm{~T}\). If the area of the loop is \(300 \mathrm{~cm}^{2}\), find the maximum angular speed possible for this motor when connected to a source of emf providing \(170 \mathrm{~V}\).

A solenoid with 200 turns and a cross-sectional area of \(60 \mathrm{~cm}^{2}\) has a magnetic field of \(0.60 \mathrm{~T}\) along its axis. If the field is confined within the solenoid and changes at a rate of \(0.20 \mathrm{~T} / \mathrm{s}\), the magnitude of the induced potential difference in the solenoid will be a) \(0.0020 \mathrm{~V}\). b) \(0.02 \mathrm{~V}\). c) \(0.001 \mathrm{~V}\). d) \(0.24 \mathrm{~V}\).

A magnetar (magnetic neutron star) has a magnetic field near its surface of magnitude \(4.0 \cdot 10^{10} \mathrm{~T}\) a) Calculate the energy density of this magnetic field. b) The Special Theory of Relativity associates energy with any mass \(m\) at rest according to \(E_{0}=m c^{2}(\) more on this in Chapter 35 ). Find the rest mass density associated with the energy density of part (a).

A rectangular wire loop (dimensions of \(h=15.0 \mathrm{~cm}\) and \(w=8.00 \mathrm{~cm}\) ) with resistance \(R=5.00 \Omega\) is mounted on a door. The Earth's magnetic field, \(B_{\mathrm{E}}=2.6 \cdot 10^{-5} \mathrm{~T}\), is uniform and perpendicular to the surface of the closed door (the surface is in the \(x z\) -plane). At time \(t=0,\) the door is opened (right edge moves toward the \(y\) -axis) at a constant rate, with an opening angle of \(\theta(t)=\omega t,\) where \(\omega=3.5 \mathrm{rad} / \mathrm{s}\) Calculate the direction and the magnitude of the current induced in the loop, \(i(t=0.200 \mathrm{~s})\).

A steel cylinder with radius \(2.5 \mathrm{~cm}\) and length \(10.0 \mathrm{~cm}\) rolls without slipping down a ramp that is inclined at \(15^{\circ}\) above the horizontal and has a length (along the ramp) of \(3.0 \mathrm{~m} .\) What is the induced potential difference between the ends of the cylinder at the bottom of the ramp, if the surface of the ramp points along magnetic north?

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