Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 28: Problem 62

You are standing at a spot where the magnetic field of the Earth is horizontal, points due northward, and has magnitude \(40.0 \mu \mathrm{T}\). Directly above your head, at a height of \(12.0 \mathrm{~m},\) a long, horizontal cable carries a steady \(\mathrm{DC}\) current of 500.0 A due northward. Calculate the angle \(\theta\) by which your magnetic compass needle is deflected from true magnetic north by the effect of the cable. Don't forget the sign of \(\theta-\) is the deflection eastward or westward?

Short Answer

Expert verified
Given: Magnetic field of the Earth: \(40.0 \, \mu T\) Height of the cable above the ground: \(12.0 \, m\) Current carried by the cable: \(500.0 A\) Following the steps outlined in the solution, carry out the calculations and find the angle of deflection, which is: \(\theta = -22.7°\) The compass needle is deflected by an angle of \(-22.7°\) westward from the true magnetic north due to the effect of the horizontal cable carrying the steady DC current.

Step by step solution

01

Calculate the magnetic field due to the wire

First, we need to calculate the magnetic field (\(B_{wire}\)) produced by the wire at the position of the person using the formula for the magnetic field produced by a long straight current-carrying wire. The formula is given by: \(B_{wire} = \frac{\mu_0 I}{2\pi d}\) Where \(B_{wire}\) is the magnetic field produced by the wire, \(\mu_0\) is the permeability of free space (\(4\pi \times 10^{-7} \, Tm/A\)), \(I\) is the current in the wire (500.0 A), and \(d\) is the distance from the wire to the person (12.0 m). Now, let's plug in the values and calculate the magnetic field due to the wire. \(B_{wire} = \frac{4\pi \times 10^{-7} \, Tm/A * 500.0 \, A}{2\pi * 12.0 \, m} = 1.67 \times 10^{-5} \, T\)
02

Find the net magnetic field

Next, let's find the net magnetic field (\(B_{net}\)) by adding the Earth's magnetic field (\(B_{earth}\)) which is given as \(40.0 \, \mu T\) (horizontal and northward) and the magnetic field due to the wire (\(B_{wire}\)), which is horizontal and points westward. Since the Earth's magnetic field and the magnetic field due to the wire are perpendicular to each other, we can use the Pythagorean theorem to find the magnitude of the net magnetic field. \(B_{net} = \sqrt{B_{earth}^2 + B_{wire}^2} = \sqrt{(40.0 \times 10^{-6}\, T)^2 + (1.67 \times 10^{-5}\, T)^2} = 4.32 \times 10^{-5} \, T\)
03

Calculate the angle of deflection

Now that we have the net magnetic field, we can find the angle \(\theta\) of deflection by considering the ratio of the magnetic field due to the wire (\(B_{wire}\)) and the Earth's magnetic field (\(B_{earth}\)). We can use the tangent function and calculate the angle as follows: \(\tan{\theta} = \frac{B_{wire}}{B_{earth}}\) \(\theta = \arctan{\frac{B_{wire}}{B_{earth}}} = \arctan{\frac{1.67 \times 10^{-5}\, T}{40.0 \times 10^{-6} \, T}} = 22.7°\) Since the magnetic field due to the wire points westward, the compass needle is deflected by \(22.7°\) westward (which is negative in this context). So, \(\theta = -22.7°\).

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.

Start your free trial

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

28.39 The figure shows a cross section across the diameter of a long, solid, cylindrical conductor. The radius of the cylinder is \(R=10.0 \mathrm{~cm} .\) A current of \(1.35 \mathrm{~A}\) is uniformly distributed through the conductor and is flowing out of the page. Calculate the direction and the magnitude of the magnetic field at positions \(r_{\mathrm{a}}=0.0 \mathrm{~cm}\) \(r_{\mathrm{b}}=4.00 \mathrm{~cm}, r_{\mathrm{c}}=10.00 \mathrm{~cm}\) and \(r_{d}=16.0 \mathrm{~cm} .\)

A square loop, with sides of length \(L\), carries current i. Find the magnitude of the magnetic field from the loop at the center of the loop, as a function of \(i\) and \(L\).

Suppose that the magnetic field of the Earth were due to a single current moving in a circle of radius \(2.00 \cdot 10^{3} \mathrm{~km}\) through the Earth's molten core. The strength of the Earth's magnetic field on the surface near a magnetic pole is about \(6.00 \cdot 10^{-5} \mathrm{~T}\). About how large a current would be required to produce such a field?

Many electrical applications use twisted-pair cables in which the ground and signal wires spiral about each other. Why?

A current element produces a magnetic field in the region surrounding it. At any point in space, the magnetic field produced by this current element points in a direction that is a) radial from the current element to the point in space. b) parallel to the current element. c) perpendicular to the current element and to the radial direction.
See all solutions

Recommended explanations on Physics Textbooks

Engineering Physics

Read Explanation

Oscillations

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Physics of Motion

Read Explanation

Electromagnetism

Read Explanation

Electric Charge Field and Potential

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.

Sign-up for free