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.

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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.

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Most popular questions from this chapter

A particle with mass \(m\) and charge \(q\) is moving within both an electric field and a magnetic field, \(\vec{E}\) and \(\vec{B}\). The particle has velocity \(\vec{v},\) momentum \(\vec{p}\), and kinetic energy, \(K\). Find general expressions for \(d \vec{p} / d t\) and \(d K / d t\), in terms of these seven quantities.

A proton moving with a speed of \(4.0 \cdot 10^{5} \mathrm{~m} / \mathrm{s}\) in the positive \(y\) -direction enters a uniform magnetic field of \(0.40 \mathrm{~T}\) pointing in the positive \(x\) -direction. Calculate the magnitude of the force on the proton.

An electron is moving with a constant velocity. When it enters an electric field that is perpendicular to its velocity, the electron will follow a ________ trajectory. When the electron enters a magnetic field that is perpendicular to its velocity, it will follow a ________ trajectory.

An electron moves in a circular trajectory with radius \(r_{\mathrm{i}}\) in a constant magnetic field. What is the final radius of the trajectory when the magnetic field is doubled? a) \(\frac{r_{i}}{4}\) b) \(\frac{r_{i}}{2}\) c) \(r_{i}\) d) \(2 r_{i}\) e) \(4 r_{\mathrm{i}}\)

A copper wire with density \(\rho=8960 \mathrm{~kg} / \mathrm{m}^{3}\) is formed into a circular loop of radius \(50.0 \mathrm{~cm} .\) The cross-sectional area of the wire is \(1.00 \cdot 10^{-5} \mathrm{~m}^{2},\) and a potential difference of \(0.012 \mathrm{~V}\) is applied to the wire. What is the maximum angular acceleration of the loop when it is placed in a magnetic field of magnitude \(0.25 \mathrm{~T}\) ? The loop rotates about an axis through a diameter.

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