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Chapter 20: Problem 28

A horizontal rod 0.200 \(\mathrm{m}\) long carries a current through a uniform horizontal magnetic field of magnitude 0.067 T that points perpendicular to the rod. If the magnetic force on this rod is measured to be \(0.13 \mathrm{N},\) what is the current flowing through the rod?

Short Answer

Expert verified
The current flowing through the rod is 9.70 A.

Step by step solution

01

Identify the Known Values

First, extract and list the given values from the problem. You have the length of the rod, \( L = 0.200 \, \text{m} \), the magnetic field, \( B = 0.067 \, \text{T} \), and the force on the rod, \( F = 0.13 \, \text{N} \).
02

Recall the Magnetic Force Formula

The formula to calculate the force on a current-carrying conductor in a magnetic field is given by: \[ F = IBL \sin(\theta) \] where \( F \) is the magnetic force, \( I \) is the current, \( B \) is the magnetic field strength, \( L \) is the length of the conductor, and \( \theta \) is the angle between the magnetic field and the current direction.
03

Determine the Angle

Since the magnetic field is perpendicular to the rod, the angle \( \theta \) is \( 90^\circ \), and \( \sin(90^\circ) = 1 \). Substitute this into the formula to simplify it to: \[ F = IBL \]
04

Solve for the Current

Rearrange the simplified formula \( F = IBL \) to solve for the current \( I \): \[ I = \frac{F}{BL} \]
05

Substitute the Values

Now, substitute the known values into the formula: \[ I = \frac{0.13}{0.067 \times 0.200} \]
06

Calculate the Current

Calculate the expression from the previous step: \[ I = \frac{0.13}{0.0134} = 9.70 \, \text{A} \] The current flowing through the rod is \( 9.70 \, \text{A} \).

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Key Concepts

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

Current
The concept of current is quite vital in understanding magnetic force interactions. In simple terms, electric current is the flow of electric charge, typically described rather like water flowing through a pipe. In a metal conductor, current is formed by the movement of electrons. This process allows for the transfer of energy across the wire.
  • Current is measured in amperes (A).
  • It has a direction, traditionally from positive to negative.
  • The current flowing through a specific area determines the magnetic force exerted on a conductor.
Understanding these basics helps to appreciate how the electrons in the rod of the exercise create such a force when interacting with the magnetic field.
Magnetic Field
A magnetic field is an invisible area around a magnetic material or an electric current that behaves as a magnet. In the context of this exercise, the magnetic field is uniform, meaning its strength and direction are constant across the area.
Magnetic fields can be thought of as lines emanating from the north to the south pole of a magnet.
  • A magnetic field is measured in teslas (T).
  • It has both strength and direction.
  • Magnetic fields can exert forces on moving charges, like in the current-carrying rod.
The direction of the magnetic field and the current determine the force acting on the conductor, as observed in the rod scenario.
Current-Carrying Conductor
A current-carrying conductor is essentially any object that allows electric current to pass through it. In the exercise, the rod acts as a conductor.
When a conductor with current is placed in a magnetic field, it experiences a force called the magnetic force. This phenomenon is a good demonstration of the interaction between electricity and magnetism.
  • The force is proportional to the length of the conductor, the current it carries, and the strength of the magnetic field.
  • If the current flows perpendicular to the magnetic field, as in the exercise situation, the force is maximized.
  • This force can be calculated using the formula: \( F = IBL \).
By applying this formula, students can determine the current flowing through a conductor when the force, length, and magnetic field are known.

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Most popular questions from this chapter

\(\cdot\) A solenoid contains 750 coils of very thin wire evenly wrapped over a length of 15.0 \(\mathrm{cm} .\) Each coil is 0.800 \(\mathrm{cm}\) in diameter. If this solenoid carries a current of \(7.00 \mathrm{A},\) what is the magnetic field at its center?

\(\bullet\) Atom smashers! A cyclotron particle accelerator (sometimes called an "atom smasher" in the popular press) is a device for accelerating charged particles, such as electrons and protons, to speeds close to the speed of light. The basic design is quite simple. The particle is bent in a circular path by a uniform magnetic field. An electric field is pulsed periodically to increase the speed of the particle. The charged particle (or ion) of mass \(m\) and charge \(q\) is introduced into the cyclotron so that it is moving perpendicular to a uniform magnetic field \(\vec{B}\) (a) Starting with the radius of the circular path of a charge moving in a uniform magnetic field, show that the time \(T\) for this particle to make one complete circle is \(T=\frac{2 \pi m}{|q| B}\) . (Hint: You can express the speed \(v\) in terms of \(R\) and \(T\) because the particle travels through one circumference of the circle in time \(T\) . (b) Which would take longer to complete one circle, an ion moving in a large circle or one moving in a small circle? Explain.

A deuteron particle (the nucleus of an isotope of hydrogen consisting of one proton and one neutron and having a mass of \(3.34 \times 10^{-27} \mathrm{kg}\) ) moving horizontally enters a uniform, vertical, 0.500 T magnetic field and follows a circular arc of radius 55.6 \(\mathrm{cm} .\) (a) How fast was this deuteron moving just before it entered the magnetic field and just after it came out of the field? (b) What would be the radius of the arc followed by a proton that entered the field with the same velocity as the deuteron?

A particle with a charge of \(-2.50 \times 10^{-8} \mathrm{C}\) is moving with an instantaneous velocity of magnitude 40.0 \(\mathrm{km} / \mathrm{s}\) in the \(x y-\) plane at an angle of \(50^{\circ}\) counterclockwise from the \(+x\) axis. What are the magnitude and direction of the force exerted on this particle by a magnetic field with magnitude 2.00 T in the (a) \(-x\) direction, and (b) \(+z\) direction?

An alpha particle (a He nucleus, containing two protons and two neutrons and having a mass of \(6.64 \times 10^{-27}\) kg traveling horizontally at 35.6 \(\mathrm{km} / \mathrm{s}\) enters a uniform, vertical, 1.10 \(\mathrm{T}\) magnetic field. (a) What is the diameter of the path followed by this alpha particle? (b) What effect does the magnetic field have on the speed of the particle? (c) What are the magnitude and direction of the acceleration of the alpha particle while it is in the magnetic field? (d) Explain why the speed of the particle does not change even though an unbalanced external force acts on it.
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