/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 27 A horizontal rod \(0.200 \mathrm... [FREE SOLUTION] | 91Ó°ÊÓ

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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 approximately 9.70 A.

Step by step solution

01

Understanding the Given Information

We are given a rod of length 0.200 m carrying a current through a magnetic field of magnitude 0.067 T. The force on the rod due to this field is 0.13 N. We need to find the current flowing through the rod.
02

Using Lorentz Force Equation

The magnetic force on a current-carrying conductor is given by the equation: \[ F = B imes I imes L imes ext{sin}( heta) \]where \(F\) is the magnetic force, \(B\) is the magnetic field, \(I\) is the current, \(L\) is the length of the rod, and \(\theta\) is the angle between the current direction and the magnetic field. Since the field is perpendicular to the rod, \(\theta = 90^\circ\) and \(\text{sin}(90^\circ) = 1\). So the equation becomes:\[ F = B imes I imes L \]
03

Rearranging the Formula to Solve for Current

Rearrange the formula \(F = B \cdot I \cdot L\) to solve for \(I\):\[ I = \frac{F}{B \times L} \]
04

Substituting the Given Values

Now substitute the given values into the rearranged formula:\[ I = \frac{0.13 \, \text{N}}{0.067 \, \text{T} \times 0.200 \, \text{m}} \]
05

Calculating the Current

Compute the expression:\[ I = \frac{0.13}{0.067 \times 0.200} \approx 9.70 \, \text{A} \]Therefore, the current flowing through the rod is approximately 9.70 A.

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Key Concepts

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

Magnetic Force
The concept of magnetic force is crucial for understanding how magnets and electricity interact. A magnetic force arises when a charged particle moves through a magnetic field. In practical terms, you often see this force at work in motors and generators. Magnetic forces can be calculated using the formula:
  • \[ F = B \times I \times L \times \sin(\theta) \]
Here, \(F\) represents the magnetic force, \(B\) is the magnetic field strength, \(I\) is the current, \(L\) is the length of the conductor, and \(\theta\) is the angle between the magnetic field and the current direction.
In many cases, including our exercise, the magnetic field is perpendicular to the conductor, making \(\theta = 90^\circ\) and \(\sin(90^\circ) = 1\). This simplifies the equation to:
  • \[ F = B \times I \times L \]
The magnetic force then purely depends on the field strength, current, and length of the conductor. This allows us to solve for the unknown, whether it's the force, current, or another variable.
Current Calculation
Calculating current is an essential skill in electromagnetism, especially when dealing with magnetic fields. To find the current flowing through a conductor when subjected to a magnetic field, the rearranged form of the magnetic force equation is used:
  • \[ I = \frac{F}{B \times L} \]
In this context:
  • \(F\) is the magnetic force acting on the conductor.
  • \(B\) is the magnetic field strength.
  • \(L\) is the length of the conductor.
For the exercise presented, substituting in the known values (force = 0.13 N, magnetic field = 0.067 T, and length = 0.200 m) into the formula gives us:
  • \[ I = \frac{0.13\, \text{N}}{0.067\, \text{T} \times 0.200\, \text{m}} \]
  • This calculation results in a current \(I \approx 9.70\, \text{A}\).
Understanding how to manipulate and apply this formula is vital for solving numerous real-world and theoretical physics problems.
Magnetic Field
The magnetic field is a vector field that surrounds magnets and moving charges. It's what enables magnets to attract or repel each other. This invisible field is depicted by lines of force, with the direction indicating the path a positive test charge would follow when placed in the field.
Key attributes of magnetic fields include:
  • Magnetic field strength, denoted by \(B\), measured in teslas (T). It reflects the field's intensity or how strong it is.
  • Direction, which is from the north pole to the south pole within the magnetic field lines.
  • Properties can vary across space, affecting how magnetic fields interact with different materials and charges.
Understanding magnetic fields is crucial for explaining phenomena such as the Earth's magnetosphere, the operation of compasses, and even complex systems like MRI machines, where they manipulate fields to create images of the human body.
In the context of the exercise, the uniform magnetic field acts perpendicular to the rod. This orientation is important as it maximizes the force impact, hence simplifying our calculations by using \( \theta = 90^\circ \) and "sin" value of 1.

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Most popular questions from this chapter

A current-carrying wire of length \(0.15 \mathrm{~m}\) is in a perpendicular magnetic field. The magnetic force on the wire is measured as a function of the current through the wire. The resulting data are given in the table. $$\begin{array}{ll}\hline \text { Current (A) } & \text { Force (N) } \\\\\hline 0.5 & 3.0 \times 10^{-3} \\\0.8 & 9.6 \times 10^{-3} \\\1.4 & 1.7 \times 10^{-2} \\\1.9 & 2.3 \times 10^{-2} \\\\\hline\end{array}$$ Make a plot of the force as a function of the current. Using a "best fit" to the data, determine the magnitude of the magnetic field.

Two high-current transmission lines carry currents of \(25 \mathrm{~A}\) and 75 A in the same direction and are suspended parallel to each other \(35 \mathrm{~cm}\) apart. If the vertical posts supporting these wires divide the lines into straight \(15 \mathrm{~m}\) segments, what magnetic force does each segment exert on the other? Is this force attractive or repulsive?

A rectangular \(10.0 \mathrm{cmby} 20.0 \mathrm{~cm}\) circuit carrying an 8.00 A current is oriented with its plane parallel to a uniform \(0.750 \mathrm{~T}\) magnetic field (Figure 20.60 ). (a) Find the magnitude and direction of the magnetic force on each segment \((a b, b c, c d,\) and da) of this circuit. Illustrate your answers with clear diagrams. (b) Find the magnitude of the net force on the entire circuit.

As a new electrical technician, you are designing a large solenoid to produce a uniform 0.150 T magnetic field near its center. You have enough wire for 4000 circular turns, and the solenoid must be \(1.40 \mathrm{~m}\) long and \(2.00 \mathrm{~cm}\) in diameter. What current will you need to produce the necessary field?

A closely wound, circular coil with radius \(2.40 \mathrm{~cm}\) has 800 turns. What must the current in the coil be if the magnetic field at the center of the coil is \(0.0580 \mathrm{~T} ?\)

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