Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A straight wire carrying a current of 3.41 A is placed at an angle of \(10.0^{\circ}\) to the horizontal between the pole tips of a magnet producing a field of \(0.220 \mathrm{~T}\) upward. The poles tips each have a \(10.0 \mathrm{~cm}\) diameter. The magnetic force causes the wire to move out of the space between the poles. What is the magnitude of that force?

Short Answer

Expert verified
Answer: The magnitude of the force on the wire is approximately 0.129 N.

Step by step solution

01

Convert the length of wire to meters.

Given the diameter of the pole tips as 10 cm, we assume the length of wire under the influence of the magnetic field (L) is also 10 cm. To convert it into meters: \( L = \frac{10}{100} = 0.1 \: m\)
02

Convert the angle to radians.

The angle is given in degrees, but the trigonometric functions (__sin__ in this case) need the angle in radians. Convert the angle from degrees to radians using the formula: \(\textrm{Angle in radians} = \frac{\textrm{Angle in degrees} \cdot \pi}{180}\) So, \(\theta = \frac{10 \cdot \pi}{180} = \frac{\pi}{18}\mathrm{~radians}\).
03

Calculate the magnetic force on the wire.

Now, using the formula \(F = I \cdot L \cdot B \cdot sin(\theta)\), we can calculate the magnitude of the magnetic force on the wire: \(F = (3.41 \: A) \cdot (0.1 \: m) \cdot (0.220 \: T) \cdot sin(\frac{\pi}{18})\) Using a calculator to perform the arithmetic, we get: \(F \approx 0.129 \: N\) So the magnitude of the force on the wire is approximately 0.129 N.

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Magnetic Field Strength
Magnetic field strength, often symbolized as B, is a quantitative expression of the intensity of a magnetic field at a particular location. Measured in teslas (T), it represents how strong the magnetic influence is that can exert a force on moving charges or current-carrying conductors.

In the given exercise, the magnetic field strength is provided as 0.220 T, indicating a relatively strong field capable of influencing nearby magnetic materials or electrical currents. When a wire with an electric current is placed within a magnetic field, this strength will directly impact the magnitude of the force that acts on it.
Current-Carrying Conductor
A current-carrying conductor, such as a wire, is a material that allows electric charge to flow through it. When electric current passes through a conductor and it is situated within a magnetic field, it experiences a force—this is the fundamental principle behind electric motors and generators.

In our scenario, the wire carries a current of 3.41 A, and due to this current, it will interact with the magnetic field between the poles of the magnet. The force on the wire depends on the current magnitude, the length of the wire within the magnetic field, and the orientation of the wire relative to the field.
Lorentz Force
The Lorentz force is the combined electric and magnetic force on a point charge due to electromagnetic fields. For a current-carrying wire, this force can be explained as the force exerted by a magnetic field B on a wire of length L carrying a current I. The magnitude of the force is given by the formula
\[ F = I \cdot L \cdot B \cdot \sin(\theta) \]

In this formula, \( \theta \) is the angle between the wire and the direction of the magnetic field. The sine function represents the component of the force perpendicular to the current flow, which is solely responsible for the motor effect – the physical movement of the wire in this context. The calculated force in our exercise is a direct application of the Lorentz force equation.
Angle Conversion Radians Degrees
Angles can be measured in both degrees and radians. While degrees are often more intuitive for people to understand, radians are the standard unit of angular measure used in mathematics, especially in trigonometry and calculus.

The conversion between degrees and radians is vital in solving physics problems involving angular measurements. One full revolution around a circle corresponds to \(360^\circ\) or \(2\pi\) radians. Therefore, to convert degrees to radians, we multiply by \(\frac{\pi}{180}\).
\[ \text{Radians} = \text{Degrees} \cdot \frac{\pi}{180} \]

Applying this to the exercise, the \(10.0^\circ\) angle is converted to radians to use in the Lorentz force formula, resulting in \(\theta = \frac{\pi}{18}\) radians.

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 helium leak detector uses a mass spectrometer to detect tiny leaks in a vacuum chamber. The chamber is evacuated with a vacuum pump and then sprayed with helium gas on the outside. If there is any leak, the helium molecules pass through the leak and into the chamber, whose volume is sampled by the leak detector. In the spectrometer, helium ions are accelerated and released into a tube, where their motion is perpendicular to an applied magnetic field, \(\vec{B},\) and they follow a circular orbit of radius \(r\) and then hit a detector. Estimate the velocity required if the orbital radius of the ions is to be no more than \(5 \mathrm{~cm},\) the magnetic field is \(0.15 \mathrm{~T}\) and the mass of a helium- 4 atom is about \(6.6 \cdot 10^{-27} \mathrm{~kg}\). Assume that each ion is singly ionized (has one electron less than the neutral atom). By what factor does the required velocity change if helium- 3 atoms, which have about \(\frac{3}{4}\) as much mass as helium- 4 atoms, are used?

The work done by the magnetic field on a charged particle in motion in a cyclotron is zero. How, then, can a cyclotron be used as a particle accelerator, and what essential feature of the particle's motion makes it possible? A blue problem number indicates a worked-out solution is available in the Student Solutions Manual. One \(\bullet\) and two \(\bullet\) indicate increasing level of problem difficulty.

A simple galvanometer is made from a coil that consists of \(N\) loops of wire of area \(A .\) The coil is attached to a mass, \(M\), by a light rigid rod of length \(L\). With no current in the coil, the mass hangs straight down, and the coil lies in a horizontal plane. The coil is in a uniform magnetic field of magnitude \(B\) that is oriented horizontally. Calculate the angle from the vertical of the rigid rod as a function of the current, \(i\), in the coil.

An electron with energy equal to \(4.00 \cdot 10^{2} \mathrm{eV}\) and an electron with energy equal to \(2.00 \cdot 10^{2} \mathrm{eV}\) are trapped in a uniform magnetic field and move in circular paths in a plane perpendicular to the magnetic field. What is the ratio of the radii of their orbits?

A coil consists of 120 circular loops of wire of radius \(4.8 \mathrm{~cm} .\) A current of 0.49 A runs through the coil, which is oriented vertically and is free to rotate about a vertical axis (parallel to the \(z\) -axis). It experiences a uniform horizontal magnetic field in the positive \(x\) -direction. When the coil is oriented parallel to the \(x\) -axis, a force of \(1.2 \mathrm{~N}\) applied to the edge of the coil in the positive \(y\) -direction can keep it from rotating. Calculate the strength of the magnetic field.

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