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A very long, straight wire carries a current of 0.12 A. This wire is tangent to a single-turn, circular wire loop that also carries a current. The directions of the currents are such that the net magnetic field at the center of the loop is zero. Both wires are insulated and have diameters that can be neglected. How much current is there in the loop?

Short Answer

Expert verified
The current in the loop is 0.12 A.

Step by step solution

01

Understand the Problem

We need to find the amount of current in a circular loop when its magnetic field at the center cancels out the magnetic field of a nearby long straight wire.
02

Identify the Magnetic Field Contributions

The magnetic field at the center of the loop has two contributions: one from the straight wire and one from the loop itself. The total magnetic field is zero, which means these two contributions must be equal in magnitude but opposite in direction.
03

Calculate the Magnetic Field due to the Straight Wire

The magnetic field due to a long straight wire at a distance \( r \) from the wire is given by the formula:\[B_{ ext{wire}} = \frac{ ext{μ}_0 I_{ ext{wire}}}{2\pi r}\]where \( I_{ ext{wire}} = 0.12 \text{ A} \) and \( r \) is the radius of the loop since the wire is tangent to the loop.
04

Calculate the Magnetic Field due to the Loop

The magnetic field at the center of a circular loop is given by the formula:\[B_{ ext{loop}} = \frac{ ext{μ}_0 I_{ ext{loop}}}{2R}\]where \( I_{ ext{loop}} \) is the unknown current through the loop, and \( R \) is the radius of the loop.
05

Set the Magnetic Fields Equal for Cancellation

Since the fields must cancel each other, we set the magnitudes equal:\[\frac{ ext{μ}_0 I_{ ext{wire}}}{2\pi r} = \frac{ ext{μ}_0 I_{ ext{loop}}}{2R}\]
06

Simplify and Solve for the Current in the Loop

Simplifying the equation by canceling out \( ext{μ}_0 \) and rearranging the terms gives:\[I_{ ext{loop}} = \frac{R}{r} \cdot I_{ ext{wire}}\]Since \( r = R \) (because the wire is tangent to the loop), \( I_{ ext{loop}} = I_{ ext{wire}} = 0.12 \text{ A} \).

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Key Concepts

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

Circular Wire Loop
A circular wire loop is a common component in electrical circuits. It consists of a wire bent into a circle and is often used to investigate magnetic fields. One of the interesting properties of a circular wire loop is that it produces a magnetic field at its center when an electric current flows through it.
This magnetic field is primarily influenced by the amount of current flowing and the radius of the loop. These factors work together to define the magnetic field's strength at various points.
In our exercise, the loop is crucial because it shares its proximity with a straight wire. This positioning influences how the magnetic field from the loop behaves, particularly in how it interacts and cancels out the field from the nearby straight wire.
Straight Wire
A long straight wire carrying an electric current produces a magnetic field in the space surrounding it. This field circles the wire in a direction determined by the "right-hand rule," with the thumb pointing in the direction of current flow and the fingers curling around the wire.
The strength of this magnetic field depends on the current flowing through the wire and the distance from the wire. The further away from the wire, the weaker the magnetic field becomes.
  • The formula used to calculate this magnetic field is: \[ B_{\text{wire}} = \frac{\mu_0 I_{\text{wire}}}{2\pi r} \]
  • \( I_{\text{wire}} \) is the current in the straight wire, while \( r \) represents the distance from the wire, often the radius of the loop in this context.
Understanding how a straight wire's field interacts with other fields is crucial in our problem since it needs to cancel the loop's field, creating a net sum of zero at the loop's center.
Current Cancellation
Current cancellation in magnetic fields occurs when two different magnetic fields interact and produce a net field of zero at a specific point. In the given situation, the circular wire loop and the long straight wire are set up in such a manner that their magnetic fields at the center of the loop negate each other.
For this to happen, the magnitudes of the magnetic fields must be equal, but they must be opposite in direction.
This cancellation is achieved by ensuring that the current in the loop perfectly counteracts the magnetic influence of the straight wire.
  • The basic principle here is balancing: \[ \frac{\mu_0 I_{\text{wire}}}{2\pi r} = \frac{\mu_0 I_{\text{loop}}}{2R} \]
By employing this equation, the unknown current in the loop can be calculated, ensuring that perfect cancellation is achieved to result in a net magnetic field of zero.
Magnetic Field Equation
The equations governing magnetic fields help in quantifying the interactions between currents and magnetic influences. In electromagnetic theory, several equations determine how these fields behave under different conditions.
In this exercise, the equations for magnetic fields due to a straight wire and a loop are pivotal. Those are\[ B_{\text{wire}} = \frac{\mu_0 I_{\text{wire}}}{2\pi r} \] for the long straight wire, and \[ B_{\text{loop}} = \frac{\mu_0 I_{\text{loop}}}{2R} \] for the circular loop. Here, \( \mu_0 \) is the permeability of free space, a constant that forms the foundation for calculating magnetic effects in a vacuum.
  • These formulas simplify the problem-solving process, allowing us to isolate and calculate the unknown variables, such as the current in the loop.
  • By setting these magnetic field equations equal, we apply the principle of superposition, ensuring that the fields exactly cancel.
Such mathematical tools are essential for predicting and controlling electromagnetic phenomena in practical applications and experiments.

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Most popular questions from this chapter

Hydrogen has three isotopes \({ }^{1} \mathrm{H}\left(m_{1}=m_{\mathrm{p}}\right)\) \({ }^{2} \mathrm{H}\left(m_{2} \cong 2 m_{\mathrm{p}}\right),\) and \({ }^{3} \mathrm{H}\left(m_{\mathrm{s}} \cong 3 m_{\mathrm{p}}\right),\) where \(m_{\mathrm{p}}\) is the mass of a proton \((1.67 \times\) \(10^{-27} \mathrm{kg}\) ). You and your team are tasked with constructing an isotope separator that will separate a gas of mixed hydrogen isotopes. The gas first passes through a device that atomizes it (i.e., makes sure the atoms are separate, and do not form \(\mathrm{H}_{2}\) molecules), and then ionizes the atoms (strips off their only electron) so that they have a net charge of \(+e .\) Next, the atoms (now positive ions) are accelerated between the plates of a parallel-plate capacitor with a voltage of \(2.5 \mathrm{kV}\) across it, and emerge through a hole in one of the plates as a beam. (a) What are the speeds of the three isotopes when they emerge from the capacitor plates? (b) The accelerated ions enter a region of uniform magnetic field oriented perpendicular to the velocity vector of the ion beam. What should be the magnitude of the magnetic field if you want the largest diameter of the three ion paths to be \(20.0 \mathrm{cm} ?\) (c) If you are collecting the atoms after completing a half-circle, one collector should be located \(20.0 \mathrm{cm}\) from the point where the beam enters the magnetic field. Where should the other two be located?

Electron beams are sometimes used to melt and evaporate metals in order to deposit thin metallic films on surfaces (similar to gold plating). One method is to put the material to be evaporated (called the "target") into a small tungsten cup (a crucible that has a very high melting point) and direct a beam of electrons at the target. Your team has been given the task of designing an electron-beam evaporator. The crucible is a cylinder, \(2.0 \mathrm{cm}\) in diameter and \(1.5 \mathrm{cm}\) in height, and contains a small target of pure nickel (Ni). The electrons are accelerated through a potential difference of \(V=1.20 \mathrm{kV}\), and form a beam that originates below the crucible, exactly \(3.70 \mathrm{cm}\) off its center, in the \(+x\) direction (see the drawing). (a) What is the speed of the electrons in the beam? (b) You must steer the electron beam with a magnetic field so that it curls over the lip of the cup and strikes the nickel target. Assuming that a uniform field exists above the cup (the field is zero below), what must be the radius of the beam's circular path? (c) In what direction should the field point if the beam initially approaches the cup from the \(-y\) axis? (d) What must be the magnitude of the uniform magnetic field?

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