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The magnetic field at the center of a \(1.0-\mathrm{cm}\) -diameter loop is \(2.5 \mathrm{mT}\) a. What is the current in the loop? b. A long, straight wire carries the same current you found in part a. At what distance from the wire is the magnetic field \(2.5 \mathrm{mT} ?\)

Short Answer

Expert verified
After substituting given values and calculating, in part a, the current in the loop is around 0.20 Ampere and in part b, the magnetic field at a distance of approximately 0.40 m from the wire is 2.5 mT.

Step by step solution

01

Apply Magnetic Field Formula

To find the current, we need to use the formula for the magnetic field at the center of a circular loop, given as \(B=\frac{{\mu_0}*I}{2*R}\), \(\mu_0\) is the permeability of free space which is \(4*\pi*10^{-7} T*m/A\), \(I\) is the current, \(R\) is the radius of the loop and \(B\) is the magnetic field. After rearranging the formula, the current can be found as \(I=\frac{{2*B*R}}{\mu_0}\).
02

Substitute Values into the Formula

Substitute \(B=2.5 mT=2.5*10^{-3} T\), \(R=1.0 cm=0.5*10^{-2} m\) (radius is half the diameter) and \(\mu_0=4*\pi*10^{-7} T*m/A\) into the formula. Therefore, \(I=\frac{{2*2.5*10^-3*0.5*10^-2}}{4*\pi*10^{-7}} A\) gives the current in the loop.
03

Apply Ampere’s Law

In the second part, Ampere’s law for magnetic field due to a long, straight wire is needed, which is given by \(B=\frac{{\mu_0*I}}{2*\pi*r}\), where \(r\) is the distance from the wire. Rearranging formula for \(r\) we get \(r=\frac{{\mu_0*I}}{2*\pi*B}\). Substitute the current obtained from Step 2 into this equation to get the distance.
04

Substitute Values into the Formula

Substitute values of \(I\), \(B=2.5*10^{-3} T\) and \(\mu_0=4*\pi*10^{-7} T*m/A\) obtained from the previous steps to solve for \(r\). Thus, the calculation \(r=\frac{{\mu_0*I}}{2*\pi*2.5*10^{-3}} m\) will yield the required distance from the wire where the magnetic field strength is identical to that in the center of the loop.

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Key Concepts

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

Ampere’s Law
Ampere's Law is a fundamental principle connecting the magnetic field and the electric current that produces it. This relationship can be expressed with the formula \(B = \frac{{\mu_0 I}}{{2 \pi r}}\), where \(B\) is the magnetic field, \(\mu_0\) represents the permeability of free space, \(I\) denotes the current, and \(r\) is the distance from the wire.

In the context of the exercise, Ampere’s Law is used to determine the magnetic field a certain distance away from a long, straight wire carrying an electric current. By manipulating this formula, you can either find the current given the magnetic field or the distance from the wire for a specific magnetic field strength, exemplified in part b of the exercise.
Magnetic Field Due to a Circular Loop
The magnetic field at the center of a circular loop is created by a current flowing through the wire. The strength of this magnetic field can be calculated using the formula \(B = \frac{{\mu_0 I}}{{2 R}}\), with \(B\) being the magnetic field, \(\mu_0\) the permeability of free space, \(I\) the current, and \(R\) the radius of the loop.

Right at the center of the loop, the magnetic field is most uniform and directly proportional to the current in the loop. This formula showcases how a current creates a magnetic field and forms the basis for understanding electromagnetism. In the given problem, it helps us to calculate the current needed to generate a specified magnetic field strength in the center of the loop.
Permeability of Free Space
The permeability of free space, often denoted by \(\mu_0\), is a constant that represents the measure of resistance encountered when forming a magnetic field in a classical vacuum. Its value is approximately \(4 \pi \times 10^{-7} \, T \cdot m / A\) (tesla meter per ampere).

This constant is crucial in the calculations of magnetic fields, as it appears in the formulas for both Ampere’s Law and the magnetic field due to a circular loop. It links magnetic and electric quantities and is a fundamental physical constant. In the exercises, we use \(\mu_0\) to calculate the current that generates the observed magnetic field and to find the distance from a wire where this magnetic field is produced.
Long Straight Wire Magnetic Field
A long straight wire carrying an electric current generates a magnetic field around it. The strength of this field at any point can be described by the formula derived from Ampere's Law: \(B = \frac{{\mu_0 I}}{{2 \pi r}}\), where \(B\) is the magnetic field, \(\mu_0\) is the permeability of free space, \(I\) is the current in the wire, and \(r\) is the radial distance from the wire.

This formula tells us that the magnetic field decreases inversely with distance from the wire, meaning the farther you are from the wire, the weaker the magnetic field becomes. Applying this concept in the exercise leads to finding out how far we must be from the wire for the magnetic field to match that of the center of the circular loop, which was calculated to be 2.5 mT.

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Most popular questions from this chapter

Two concentric current loops lie in the same plane. The smaller loop has a radius of \(3.0 \mathrm{cm}\) and a current of \(12 \mathrm{A}\). The bigger loop has a current of 20 A. The magnetic field at the center of the loops is found to be zero. What is the radius of the bigger loop?

A solenoid used to produce magnetic fields for research purposes is \(2.0 \mathrm{m}\) long, with an inner radius of \(30 \mathrm{cm}\) and 1000 turns of wire. When running, the solenoid produces a field of \(1.0 \mathrm{T}\) in the center. Given this, how large a current does it carry?

Medical cyclotrons need efficient sources of protons to inject into their center. In one kind of ion source, hydrogen atoms (i.e., protons with one orbiting electron) are fed into a chamber where there is a strong magnetic field. Electrons in this chamber are trapped in tight orbits, which greatly increases the chance that they will collide with a hydrogen atom and ionize it. One such source uses a magnetic field of \(50 \mathrm{mT}\), and the electrons' kinetic energy is \(2.0 \mathrm{eV}\). If the electrons travel in a plane perpendicular to the field, what is the radius of their orbits?

The ocean is salty because it contains many dissolved ions. As these charged particles move with the water in strong ocean currents, they feel a force from the earth's magnetic field. Positive and negative charges are separated until an electric field develops that balances this magnetic force. This field produces measurable potential differences that can be monitored by ocean researchers. The Gulf Stream moves northward off the east coast of the United States at a speed of up to \(3.5 \mathrm{m} / \mathrm{s}\). Assume that the current flows at this maximum speed and that the earth's field is \(50 \mu \mathrm{T}\) tipped \(60^{\circ}\) below horizontal. What is the direction of the magnetic force on a singly ionized negative chlorine ion moving in this ocean current? A. East B. West C. Up D. Down

A medical cyclotron used in the production of medical isotopes accelerates protons to 6.5 MeV. The magnetic field in the cyclotron is \(1.2 \mathrm{T}\). a. What is the diameter of the largest orbit, just before the protons exit the cyclotron? b. A proton exits the cyclotron \(1.0 \mathrm{ms}\) after starting its spiral trajectory in the center of the cyclotron. How many orbits does the proton complete during this \(1.0 \mathrm{ms} ?\)

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