/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 74 If part of the magnet develops r... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If part of the magnet develops resistance and liquid helium boils away, rendering more and more of the magnet nonsuperconducting, how will this quench affect the time for the current to drop to half of its initial value? (a) The time will be shorter because the resistance will increase; (b) the time will be longer because the resistance will increase; (c) the time will be the same; (d) not enough information is given.

Short Answer

Expert verified
(a) The time will be shorter because the resistance will increase.

Step by step solution

01

Understanding the Situation

When a superconducting magnet develops resistance, and liquid helium boils away, it leads to a quench. A quench causes part or all of the magnet to lose its superconducting state, turning it into a resistive state.
02

Impact of Resistance on Current

With resistance introduced in the magnet, the current will start to experience energy loss in the form of heat, causing it to decay more quickly than when it was superconducting.
03

Relation between Resistance and Time

In a resistive circuit, the current decay can be described by the formula \( I(t) = I_0 e^{-Rt/L} \), where \( R \) is the resistance, \( L \) is the inductance, and \( I_0 \) is the initial current. A higher resistance \( R \) leads to a faster decay of current.
04

Determine the Time for Current to Halve

The time to reduce the current to half its initial value, known as the half-life, is given by \( t = \frac{L}{R} \ln(2) \). An increase in resistance \( R \) results in a shorter time \( t \) for the current to halve due to the inverse relationship.
05

Choose the Correct Option

As resistance increases, the time for the current to halve decreases. Thus, the correct answer is option (a): The time will be shorter because the resistance will increase.

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 91Ó°ÊÓ!

Key Concepts

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

Resistance in Superconductors
In the realm of superconductivity, resistance is a fascinating concept because it's essentially nonexistent. Superconductors are special materials that, below a certain temperature, can conduct electricity with zero resistance. This means there's no energy loss in the form of heat, making superconductors extremely efficient conductors. However, when conditions change—such as the increase in temperature due to the loss of liquid helium—parts of a superconducting magnet may revert to a resistive state. This transition, known as a quench, introduces resistance into the circuit.
  • Superconductors have zero electrical resistance when cooled below a critical temperature.
  • A quench introduces resistance, causing the loss of superconductivity in parts of the magnet.
  • Increased resistance leads to energy loss in the form of heat, impacting the current flow.
When resistance is introduced, even gradually, the energy formerly traveling effortlessly through the superconducting pathway faces increased opposition. This leads to a portion of the energy being converted to heat, drastically affecting the behavior of the current passing through.
Current Decay in Magnetic Circuits
Current decay in magnetic circuits that once housed superconductors behaves differently due to the presence of resistance. Normally, in a superconducting circuit, the current could circulate endlessly with no decrease. However, when resistance comes into play, as it does during a quench, the current begins to follow an exponential decay pattern.
This decay is mathematically expressed using the formula \( I(t) = I_0 e^{-Rt/L} \), which shows how the current \( I(t) \) decreases over time. Here, \( R \) represents the resistance, \( L \) is the inductance of the circuit, and \( I_0 \) is the initial current. A larger resistance results in a steeper decline. Thus, the current decays faster, leading to a shorter time for the current to reach half of its original value.
  • Exponential decay occurs when resistance is present in the circuit.
  • Higher resistance leads to faster decay, impacting how quickly the current reduces.
  • The half-life of the current is shorter with increased resistance, calculated as \( t = \frac{L}{R} \ln(2) \).
Through this understanding, it's clear why, when a superconducting magnet transitions to having resistance, the time for the current to halve decreases.
Superconductivity and Liquid Helium
To maintain the superconducting state, low temperatures are crucial, usually achievable with coolants like liquid helium. Liquid helium, particularly at temperatures close to absolute zero, provides the cold environment necessary for materials to transition into the superconducting state where they can carry electric current without resistance.
However, if the liquid helium starts to boil away, the temperature of the superconducting magnet rises. This increase can trigger a quench, causing parts or all of the magnet to lose its superconducting properties.
  • Liquid helium cools systems to extremely low temperatures essential for superconductivity.
  • The boiling away of liquid helium increases the temperature, risking a quench.
  • Without adequate cooling, superconductors revert to normal resistive behavior, affecting performance.
  • When superconductors return to a resistive state due to inadequate cooling, not only does the efficiency drop, but managing power loss and ensuring system stability becomes crucial.

    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 coil has 400 turns and self-inductance 7.50 mH. The current in the coil varies with time according to \(i = (680 \, \mathrm{mA}) \mathrm{cos} (\pi{t}/0.0250 \, \mathrm{s})\). (a) What is the maximum emf induced in the coil? (b) What is the maximum average flux through each turn of the coil? (c) At \(t = 0.0180\) s, what is the magnitude of the induced emf?

    It is proposed to store \(1.00 \, \mathrm{kW} \cdot \mathrm{h} = 3.60 \times 10^6\) J of electrical energy in a uniform magnetic field with magnitude 0.600 T. (a) What volume (in vacuum) must the magnetic field occupy to store this amount of energy? (b) If instead this amount of energy is to be stored in a volume (in vacuum) equivalent to a cube 40.0 cm on a side, what magnetic field is required?

    It has been proposed to use large inductors as energy storage devices. (a) How much electrical energy is converted to light and thermal energy by a 150-W light bulb in one day? (b) If the amount of energy calculated in part (a) is stored in an inductor in which the current is 80.0 A, what is the inductance?

    A 6.40-nF capacitor is charged to 24.0 V and then disconnected from the battery in the circuit and connected in series with a coil that has \(L =\) 0.0660 H and negligible resistance. After the circuit has been completed, there are current oscillations. (a) At an instant when the charge of the capacitor is 0.0800 \(\mu\)C, how much energy is stored in the capacitor and in the inductor, and what is the current in the inductor? (b) At the instant when the charge on the capacitor is 0.0800 \(\mu\)C, what are the voltages across the capacitor and across the inductor, and what is the rate at which current in the inductor is changing?

    When the current in a toroidal solenoid is changing at a rate of 0.0260 A/s, the magnitude of the induced emf is 12.6 mV. When the current equals 1.40 A, the average flux through each turn of the solenoid is 0.00285 Wb. How many turns does the solenoid have?

    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.