/*! 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} Q75E Question: The magnetic field at ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Q75E (page 473)

Question: The magnetic field at the surface of a long wire radius R and carrying a current I is 渭0I2蟺R . How large acurrent could a 0.1 mm diameter niobium wire carry without exceeding its 0.2 T critical field?

Short Answer

Expert verified

Answer

Current in the wire is 50A.

Step by step solution

01

Given data

Diameter of niobium wire is, 0.1mm .

Maximum magnetic field is,0.2T .

02

 Step 2: Formula of Magnetic Field produced by a Wire

given by, B=渭0I2蟺R .

Here I represents current, R represents radius of wire and渭0 represents permeability of free space.

03

Calculate the Current from the Expression of Magnetic Field produced by a Wire

The radius of the wire is, R=D2.

Substitute 0.1mm鈫�D and convert it into m .

R=D2=0.1mm2=0.05mm10-3m1mm=0.05脳10-3m

The magnetic field produced by the current is given by, B=渭0I2蟺R.

Rearrange the above equation for l .

B=渭0I2蟺R

Substitute 4蟺脳10-7T路m/A鈫�渭0,0.2T鈫�B, and 0.05脳10-3m鈫�R in the above equation.

B=渭0I2蟺RI=2蟺RB渭0=2蟺0.05脳10-3鈥�m(0.2鈥�T)4蟺脳10-7鈥�Tm/A=50鈥�A

Current in the wire is 50 A .

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 91影视!

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

Exercise 29 outlines how energy may be extracted by transferring an electron from an atom that easily loses an electron from an atom that easily loses an electron to one with a large appetite for electrons , then allowing the two to approach , forming an ionic bond.

  1. Consider separately the cases of hydrogen bonding with fluorine and sodium bonding with fluorine in each case , how close must the ions approach to reach 鈥渂reak even鈥� where the energy needed to transfer the electron between the separated atoms is balanced by the electrostatic potential energy of attraction? The ionization energy of hydrogen is 13.6 eV , that of sodium is 5.1 eV , and the electron affinity of fluorine is 3.40 Ev.
  2. Of HF and NaF , one is considered to be an ionic bond and the other a covalent bond . Which is which and Why?

Based only on the desire to limit minority carriers, why would silicon be preferable to germanium as a fabric for doped semiconductors?

Vibration-rotation spectra are rich For the CO molecule (data are given in Exercise 42), roughly how many rotational levels would there be between the ground vibrational state and the first excited vibrational state?

The left diagram in FIGURE 10.1 might represent a two atom crystal with two bands. Basing your argument on the kinetic energy inside either individual well, explain why both energies in the lower band should be roughly equal to that of the n=1 atomic state and why both energies in the upper should roughly equal that of the n=2 atomic state

Question: The diagram shows the energy bands of a tunnel diode as the potential difference is increased. In this device high impurity atom density causes the occupied donor and unoccupied acceptor levels to spread into impurity bands which overlap respectively the n-type conduction- and the p-type valence bands. In all unbiased diodes, the depletion zone between the n-type and p-type bands constitutes a potential barrier (see Section 6.2) but in the tunnel diode it is so thin that significant tunnelling occurs. The current versus voltage plot shows that unlike a normal diode significant current begins to flow as soon as there is an applied voltage鈥攂efore the bias voltage is Egap /e. It then decreases (so called negative resistance) before again increasing in the normal way. Explain this behavior.
See all solutions

Recommended explanations on Physics Textbooks

Thermodynamics

Read Explanation

Waves Physics

Read Explanation

Translational Dynamics

Read Explanation

Electricity and Magnetism

Read Explanation

Famous Physicists

Read Explanation

Energy Physics

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.