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Chapter 28: Problem 40

A wire \(1.80 \mathrm{~m}\) long carries a current of \(13.0 \mathrm{~A}\) and makes an angle of \(35.0^{\circ}\) with a uniform magnetic field of magnitude \(B=1.50\) T. Calculate the magnetic force on the wire.

Short Answer

Expert verified
The magnetic force on the wire is approximately 20.2 N.

Step by step solution

01

Understanding the Magnetic Force Formula

The magnetic force \( F \) on a current-carrying wire in a magnetic field is given by the formula \( F = I \cdot L \cdot B \cdot \sin \theta \), where \( I \) is the current, \( L \) is the length of the wire, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the wire and the magnetic field. For this problem, \( I = 13.0 \, \mathrm{A} \), \( L = 1.80 \, \mathrm{m} \), \( B = 1.50 \, \mathrm{T} \), and \( \theta = 35.0^{\circ} \).
02

Converting Angle to Radians

Since trigonometric functions in calculus typically use radians, let's convert the angle from degrees to radians using the formula \( \theta_{radians} = \theta_{degrees} \times \frac{\pi}{180} \). So, \( 35.0^{\circ} = 35.0 \times \frac{\pi}{180} \approx 0.6109 \, \mathrm{radians} \).
03

Calculating the Sine of the Angle

Using the sine function for the converted angle, \( \sin(35.0^{\circ}) \approx 0.5736 \).
04

Substitute Values into the Formula

We now substitute the values into the magnetic force formula: \( F = 13.0 \, \mathrm{A} \times 1.80 \, \mathrm{m} \times 1.50 \, \mathrm{T} \times 0.5736 \approx 20.2 \, \mathrm{N} \). This gives us the magnitude of the force acting on the wire.

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Key Concepts

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

Current-Carrying Wire
When discussing physics, specifically electromagnetism, it's essential to understand what a current-carrying wire is. A wire that conducts electricity is often part of many electronic circuits. In simple terms, a current-carrying wire is a wire through which electric current or the flow of electric charges is passing. This is typically measured in amperes (A).
In our scenario, the wire carries a current of 13.0 A. This flow of electricity can interact with magnetic fields present in its surroundings, thereby experiencing a force depending on the current itself, the magnetic field strength, and the orientation of these two entities in relation to each other.
Magnetic Field
A magnetic field is a region around a magnetic material or a moving electric charge within which the force of magnetism acts. It is often denoted by the letter 'B' and is measured in Tesla (T).
  • Magnetic fields can be visualized as lines that show the direction and strength at nearby points.
  • Moving charges, such as those in a current-carrying wire, will act upon and be affected by these fields.
In the given problem, the magnetic field strength is 1.50 T. When an electric current passes through a wire in a magnetic field, it experiences a magnetic force. The interaction between the current and the magnetic field results in this force, which can be calculated using specific formulas and depends on their perpendicularity.
Sine Function
The sine function, often seen in trigonometry and calculus, is vital in determining the effect of angles in physics problems. For instance, the effectiveness of a force acting at an angle is essentially a fraction determined by the sine of that angle.
The sine function is a component in the equation used to calculate the magnetic force on a current-carrying wire. Its value ranges from -1 to 1, describing the ratio of the length of the side opposite the angle to the hypotenuse in a right triangle. When a wire is said to make an angle of 35.0掳 with the magnetic field, the sine of this angle helps us find out how much of the magnetic field's effect is experienced by the wire.

The value of the sine for 35.0掳 is approximately 0.5736, which is crucial in defining the force through the formula.
Radians Conversion
Angles can be measured in degrees or radians, each with its application. In higher mathematics, radians are often preferred because they provide a more natural way to express angles derived from circle geometry.
  • To convert from degrees to radians, you multiply the degree value by \(\pi/180\).
In our magnetic force problem, the original angle provided is 35.0掳. To make it more suitable for trigonometric functions used in calculus, the angle needs conversion into radians.
By multiplying \ 35.0 \times \frac{\pi}{180} \ , we convert degrees to approximately 0.6109 radians. This conversion ensures the calculators and formulas we use are accurate and in the preferred unit for precise scientific calculations.

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Most popular questions from this chapter

An ion source is producing \({ }^{6} \mathrm{Li}\) ions, which have charge \(+e\) and mass \(9.99 \times 10^{-27} \mathrm{~kg}\). The ions are accelerated by a potential difference of \(10 \mathrm{kV}\) and pass horizontally into a region in which there is a uniform vertical magnetic field of magnitude \(B=1.2 \mathrm{~T}\). Calculate the strength of the smallest electric field, to be set up over the same region, that will allow the \({ }^{6} \mathrm{Li}\) ions to pass through undeflected.

At one instant, \(\vec{v}=(-2.00 \hat{\mathrm{i}}+4.00 \hat{\mathrm{j}}-6.00 \hat{\mathrm{k}}) \mathrm{m} / \mathrm{s}\) is the ve- locity of a proton in a uniform magnetic field \(\vec{B}=(2.00 \hat{\mathrm{i}}-\) \(4.00 \hat{j}+8.00 \hat{k}) \mathrm{mT}\). At that instant, what are (a) the magnetic force \(\vec{F}\) acting on the proton, in unit-vector notation, (b) the angle between \(\vec{v}\) and \(\vec{F}\), and (c) the angle between \(\vec{v}\) and \(\vec{B}\) ?

A particle with charge \(2.0 \mathrm{C}\) moves through a uniform magnetic field. At one instant the velocity of the particle is \((2.0 \hat{\mathrm{i}}+4.0 \hat{\mathrm{j}}+6.0 \hat{\mathrm{k}}) \mathrm{m} / \mathrm{s}\) and the magnetic force on the particle is \((4.0 \hat{\mathrm{i}}-20 \hat{\mathrm{j}}+12 \hat{\mathrm{k}}) \mathrm{N}\). The \(x\) and \(y\) components of the magnetic field are equal. What is \(\vec{B}\) ?

Two concentric, circu- endlar wire loops, of radii \(r_{1}=20.0 \mathrm{~cm}\) and \(r_{2}=30.0 \mathrm{~cm}\), are located in an \(x y\) plane; each carries a clockwise current of \(7.00 \mathrm{~A}\) (Fig. \(28-47\) ). (a) Find the magnitude of the net magnetic dipole moment of the system. (b) Repeat for reversed current in the inner loop.

An electron with kinetic energy \(2.5 \mathrm{keV}\) moving along the positive direction of an \(x\) axis enters a region in which a uniform electric field of magnitude \(10 \mathrm{kV} / \mathrm{m}\) is in the negative direction of the \(y\) axis. A uniform magnetic field \(\vec{B}\) is to be set up to keep the electron moving along the \(x\) axis, and the direction of \(\vec{B}\) is to be chosen to minimize the required magnitude of \(\vec{B}\). In unit-vector notation, what \(\vec{B}\) should be set up?
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