Two long, parallel wires separated by a distance, \(d\), carry currents in opposite directions. If the left-hand wire carries a current \(i / 2,\) and the right-hand wire carries a current \(i\), determine where the magnetic field is zero.

Short Answer

Expert verified
Answer: The position where the magnetic field is zero is $\frac{d}{3}$ from the left-hand wire and $\frac{2d}{3}$ from the right-hand wire, where $d$ is the distance between the two wires.

Step by step solution

01

Determine the magnetic field produced by each wire

First, let's determine the magnetic field produced by each wire using the Biot-Savart Law. The magnetic field at a distance \(x\) from a wire carrying current \(i\) is given by: \(B = \frac{\mu_0 i}{2\pi x}\) where \(\mu_0\) is the permeability of free space, equal to \(4\pi \times 10^{-7} \, Tm/A\).
02

Set up an equation for the position where the magnetic fields cancel out

We want to find the position where the magnetic fields produced by the two wires cancel out. This means that \(B_1 = B_2\), where \(B_1\) is the magnetic field due to the left-hand wire and \(B_2\) is the magnetic field due to the right-hand wire. Let \(x_1\) be the distance from the left-hand wire to the position where the magnetic fields cancel, and \(x_2\) be the distance from the right-hand wire to the same position. The magnetic fields at this position are: \(B_1 = \frac{\mu_0 (i/2)}{2\pi x_1}\) \(B_2 = \frac{\mu_0 i}{2\pi x_2}\) Since we want to find the position where \(B_1 = B_2\), we can set the expressions for \(B_1\) and \(B_2\) equal to each other: \(\frac{\mu_0 (i/2)}{2\pi x_1} = \frac{\mu_0 i}{2\pi x_2}\)
03

Solve for the ratio between \(x_1\) and \(x_2\)

Now we need to solve the equation for the ratio between \(x_1\) and \(x_2\). We can first cancel the terms that appear in both sides of the equation: \(\frac{i/2}{x_1} = \frac{i}{x_2}\) Next, let's make \(x_2\) the subject of the equation: \(x_2 = \frac{2x_1}{(i/2)}\) We can further simplify it: \(x_2 = 2x_1\) This tells us that the position at which the magnetic fields cancel out is twice as far from the right-hand wire as it is from the left-hand wire. To determine the actual position along the wire, we can use the fact that \(x_1 + x_2 = d\): \(d = x_1 + 2x_1 = 3x_1\) Thus, \(x_1 = \frac{d}{3}\) and \(x_2 = \frac{2d}{3}\).
04

Conclusion

The position where the magnetic field is zero is \(\frac{d}{3}\) from the left-hand wire and \(\frac{2d}{3}\) from the right-hand wire.

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Key Concepts

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

Biot-Savart Law
The Biot-Savart Law is a critical principle in electromagnetism, specifically concerning the magnetic effects of electric currents. It provides a mathematical model for calculating the magnetic field produced at a point in space due to a current-carrying conductor. According to this law, the magnetic field \( B \) at a point is directly proportional to the current \( i \) and inversely proportional to the distance \( x \) from the conductor.

For a straight, infinite wire, the equation is expressed as:
\[ B = \frac{\mu_0 i}{2\pi x} \]
where \( \mu_0 \) is the permeability of free space, representing how much resistance the vacuum offers against the formation of the magnetic field. This law is indispensable when dealing with problems involving magnetic fields around currents, as evident in our textbook problem where it helps determine the magnetic field produced by two parallel wires carrying electric current.
Magnetic Fields Cancellation
Magnetic fields cancellation is a phenomenon where the magnetic fields from different sources combine to produce a net field of zero at certain points in space. This occurs when the magnetic fields are equal in magnitude but opposite in direction. In our exercise, we explored this concept by looking for a point where the magnetic fields produced by two currents in opposite directions cancel each other out.

Using the Biot-Savart Law, we find that at the point of cancellation, the field contributions from both wires must satisfy the condition:\[ B_1 = B_2 \]
By setting up and solving the equations, we determined the specific locations relative to the wires where this cancellation occurs. Understanding magnetic field cancellation is crucial for designing magnetic shielding and in applications that require neutral magnetic environments, such as in certain medical or scientific instrumentation.
Permeability of free space
The permeability of free space, denoted as \( \mu_0 \) and also known as the magnetic constant, is a fundamental physical constant that represents the measure of the magnetic permeability of the classical vacuum. Essentially, it's a measure of how much magnetic field is produced in a vacuum by an electric current or a magnetic moment. The value of \( \mu_0 \) is roughly \(4\pi \times 10^{-7} \, Tm/A\) (tesla meter per ampere).

In the equations derived from the Biot-Savart Law, \( \mu_0 \) appears as a multiplier, showing how it influences the strength of the magnetic field around a current-carrying wire. It's noteworthy that \( \mu_0 \) not only plays a role in this law but also in other electromagnetic phenomena, including the inductance of coils and the operation of electromagnetic waves.

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Most popular questions from this chapter

The magnetic force cannot do work on a charged particle since the force is always perpendicular to the velocity. How then can magnets pick up nails? Consider two parallel current-carrying wires. The magnetic fields cause attractive forces between the wires, so it appears that the magnetic field due to one wire is doing work on the other wire. How is this explained? a) The magnetic force can do no work on isolated charges; this says nothing about the work it can do on charges confined in a conductor. b) Since only an electric field can do work on charges, it is actually the electric fields doing the work here. c) This apparent work is due to another type of force.

An electron has a spin magnetic moment of magnitude \(\mu=9.285 \cdot 10^{-24} \mathrm{~A} \mathrm{~m}^{2}\). Consequently, it has energy associated with its orientation in a magnetic field. If the difference between the energy of an electron that is "spin up" in a magnetic field of magnitude \(B\) and the energy of one that is "spin down" in the same magnetic field (where "up" and "down" refer to the direction of the magnetic field) is \(9.460 \cdot 10^{-25} \mathrm{~J}\), what is the field magnitude, \(B\) ?

A long, straight wire carries a current of 2.5 A. a) What is the strength of the magnetic field at a distance of \(3.9 \mathrm{~cm}\) from the wire? b) If the wire still carries \(2.5 \mathrm{~A}\), but is used to form a long solenoid with 32 turns per centimeter and a radius of \(3.9 \mathrm{~cm}\) what is the strength of the magnetic field at the center of the solenoid?

A particle detector utilizes a solenoid that has 550 turns of wire per centimeter. The wire carries a current of 22 A. A cylindrical detector that lies within the solenoid has an inner radius of \(0.80 \mathrm{~m} .\) Electron and positron beams are directed into the solenoid parallel to its axis. What is the minimum momentum perpendicular to the solenoid axis that a particle can have if it is to be able to enter the detector?

A toy airplane of mass \(0.175 \mathrm{~kg}\), with a charge of \(36 \mathrm{mC}\), is flying at a speed of \(2.8 \mathrm{~m} / \mathrm{s}\) at a height of \(17.2 \mathrm{~cm}\) above and parallel to a wire, which is carrying a 25 - A current; the airplane experiences some acceleration. Determine this acceleration.

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