/*! 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 85 Two infinitely long, thin, insul... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 85

Two infinitely long, thin, insulated, straight wires lies in the \(x-y\) plane along the \(x\) - and \(y\)-axes, respectively. Each wire carries a current \(I\), respectively, in the positive \(x\)-direction and positive \(y\)-direction. The magnetic field will be zero at all points on the straight line. (A) \(y=x\) (B) \(y=-x\) (C) \(y=x-1\) (D) \(y=-x+1\)

Short Answer

Expert verified
The magnetic field will be zero at all points on the straight line \(y=x\).

Step by step solution

01

Calculate the magnetic field due to each wire

To begin, we need to apply the Biot-Savart law for each wire and find the magnetic field. The magnetic field due to a current carrying wire at a distance r from it is given by \(B = \frac { \mu_0 I}{2 \pi r} \). Let's calculate the magnetic field at a point (x,y) in the x-y plane due to both wires. For the wire on the x-axis: Here the distance r is simply y. So, the magnetic field due to this wire will be: \[B_x = \frac{ \mu_0 I}{2 \pi y} \] And the direction of this magnetic field will be perpendicular to the x-axis. For the wire on the y-axis: In this case, the distance r is x. So, the magnetic field due to this wire will be: \[B_y = \frac{ \mu_0 I}{2 \pi x} \] And the direction of this magnetic field will be perpendicular to the y-axis. Now we have the magnetic fields due to each wire.
02

Determine where the magnetic fields cancel each other out

Now we need to find the points where the magnetic fields from each wire cancel each other out. For this to happen, the magnetic fields should have the same magnitude but opposite directions. Both the magnetic fields are orthogonal to each other. So, let's compare their magnitudes. Set \(B_x\) equal to \(B_y\): \[\frac{ \mu_0 I}{2 \pi y} = \frac{ \mu_0 I}{2 \pi x} \] We can simplify this equation by multiplying both sides by \(2 \pi\) and dividing by \(\mu_0 I\): \[ \frac{1}{y} = \frac{1}{x} \] Now, by cross-multiplying, we get: \[x = y\] Hence, the straight line where the magnetic field is zero due to these two wires is \(y = x\). The correct answer is: (A) \(y=x\)

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

A horizontal overhead powerline is at height of \(4 \mathrm{~m}\) from the ground and carries a current of \(100 \mathrm{~A}\) from east to west. The magnetic field directly below it on the ground is \(\left(\mu_{0}=4 \pi \times 10^{-7} \mathrm{Tm} A^{-1}\right)\) [2008] (A) \(2.5 \times 10^{-7} \mathrm{~T}\) southward (B) \(5 \times 10^{-6} \mathrm{~T}\) northward (C) \(5 \times 10^{-6} \mathrm{~T}\) southward (D) \(2.5 \times 10^{-7} \mathrm{~T}\) northward

Two concentric coils each of radius equal to \(2 \pi \mathrm{cm}\) are placed at right angles to each other, \(3 \mathrm{~A}\) and \(4 \mathrm{~A}\) are the currents flowing in coils, respectively. The magnetic induction in \(\mathrm{Wb} / \mathrm{m}^{2}\) at the centre of the coils will be \(\left(\mu_{0}=4 \pi \times 10^{-7} \mathrm{~Wb} / \mathrm{Am}\right)\) (A) \(12 \times 10^{-5}\) (B) \(10^{-5}\) (C) \(5 \times 10^{-5}\) (D) \(7 \times 10^{-5}\)

A particle of mass \(m\) and charge \(q\) moves with a constant velocity \(v\) along the positive \(x\) direction. It enters a region containing a uniform magnetic field \(B\) directed along the negative \(z\)-direction, extending from \(x=a\) to \(x=b .\) The minimum value of \(v\) required so that the particle can just enter the region \(x>b\) is (A) \(q b B / m\) (B) \(q(b-a) B / m\) (C) \(q a B / m\) (D) \(q(b+a) B / 2 m\)

A long solenoid has 200 turns per \(\mathrm{cm}\) and carries a current \(i\). The magnetic field at its centre is \(6.28 \times 10^{-2}\) \(\mathrm{Wb} / \mathrm{m}^{2}\) Another long solenoid has 100 turns per \(\mathrm{cm}\) and it carries a current \(\frac{i}{3} .\) The value of the magnetic field at its centre is (A) \(1.05 \times 10^{-2} \mathrm{~Wb} / \mathrm{m}^{2}\) (B) \(1.05 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^{2}\) (C) \(1.05 \times 10^{-3} \mathrm{~Wb} / \mathrm{m}^{2}\) (D) \(1.05 \times 10^{-4} \mathrm{~Wb} / \mathrm{m}^{2}\)

An electron of mass \(m\) is accelerated through a potential difference of \(V\) and then it enters a magnetic field of induction \(B\) normal to the lines. Then the radius of the circular path is (A) \(\sqrt{\frac{2 e V}{m}}\) (B) \(\sqrt{\frac{2 V m}{e B^{2}}}\) (C) \(\sqrt{\frac{2 V m}{e B}}\) (D) \(\sqrt{\frac{2 V m}{e^{2} B}}\)
See all solutions

Recommended explanations on Physics Textbooks

Solid State Physics

Read Explanation

Linear Momentum

Read Explanation

Astrophysics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Mechanics and Materials

Read Explanation

Space 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.