Figure 29-49 shows two very long straight wires (in cross section) that each carry a current of4.00A directly out of the page. Distance d1=6.00m and distance d2=4.00m. What is the magnitude of the net magnetic field at point P, which lies on a perpendicular bisector to the wires?

Short Answer

Expert verified

Magnitude of the net magnetic field at point P is 256nT.

Step by step solution

01

Given

  1. Permeability of free space, μ0=4π×10-7TmA.
  2. d1=6.00m.
  3. d2=4.00m.
  4. Current carried by wire,i=4.00A.
02

Understanding the concept

By using the concept of magnetic field due to long wire carrying current and component of magnetic field, determine the net magnetic field at point P.

Formula:

Magnetic field due to the wire at point P is

B=μ0i2πr

Here, i=current, μ0=permeability of free space,r=distance of wire.

03

Calculate the magnitude of the net magnetic field at point P.

First of all, we have to find the r.

By using Pythagoras theorem, solve as:

r=d122+d22

r=622+42

r=25r=5.00m

Now, the magnetic field is given by

B=μ0i2πr

B=4π×10-7×4.02×3.14×5.00

B=1.6×10-7T

B=160nT

The component of the magnetic field cancels with each other, so the magnetic field at point P is

Bp=2Bsinθ

Here,

sinθ=d2r=4.005.00=0.8

Bp=2×160×0.8

Bp=256nT

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

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