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

An electron has a spin magnetic moment of magnitude \(\mu=9.285 \cdot 10^{-24} \mathrm{~A} \mathrm{~m}^{2}\). Consequently, it has energy associated with its orientation in a magnetic field. If the difference between the energy of an electron that is "spin up" in a magnetic field of magnitude \(B\) and the energy of one that is "spin down" in the same magnetic field (where "up" and "down" refer to the direction of the magnetic field) is \(9.460 \cdot 10^{-25} \mathrm{~J}\), what is the field magnitude, \(B\) ?

A horizontally oriented coil of wire of radius \(5.00 \mathrm{~cm}\) and carrying a current, \(i\), is being levitated by the south pole of a vertically oriented bar magnet suspended above the center of the coil. If the magnetic field on all parts of the coil makes an angle \(\theta\) of \(45.0^{\circ}\) with the vertical, determine the magnitude and the direction of the current needed to keep the coil floating in midair. The magnitude of the magnetic field is \(B=0.0100 \mathrm{~T}\), the number of turns in the coil is \(N=10.0\), and the total coil mass is \(10.0 \mathrm{~g}\).

Can an ideal solenoid, one with no magnetic field outside the solenoid, exist? If not, does that render the derivation of the magnetic field inside the solenoid (Section 28.4) void?

A current of constant density, \(J_{0}\), flows through a very long cylindrical conducting shell with inner radius \(a\) and outer radius \(b\). What is the magnetic field in the regions \(rb\) ? Does \(B_{ab}\) for \(r=b\) ?

A 0.90 m-long solenoid has a radius of \(5.0 \mathrm{~mm} .\) When the wire carries a 0.20 - A current, the magnetic field in the solenoid is \(5.0 \mathrm{mT}\). How many turns of wire are there in the solenoid?
See all solutions

Recommended explanations on Physics Textbooks

Work Energy and Power

Read Explanation

Atoms and Radioactivity

Read Explanation

Torque and Rotational Motion

Read Explanation

Waves Physics

Read Explanation

Nuclear Physics

Read Explanation

Electricity and Magnetism

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