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In a physics lab, a student discovers that the magnitude of the magnetic field at a certain distance from a long wire is \(4.0 \mu \mathrm{T}\). If the wire carries a current of \(5.0 \mathrm{~A}\), what is the distance of the magnetic field from the wire?

Short Answer

Expert verified
The distance from the wire is 0.25 meters.

Step by step solution

01

Identify the formula to use

To find the distance from the wire where the magnetic field is given, we use the formula for the magnetic field around a long straight current-carrying wire: \( B = \frac{\mu_0 I}{2\pi r} \), where \( B \) is the magnetic field, \( I \) is the current, and \( r \) is the distance from the wire. \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \mathrm{~T} \cdot \mathrm{m/A} \)).
02

Solve for the distance \( r \)

Rearrange the formula to solve for \( r \): \( r = \frac{\mu_0 I}{2\pi B} \).
03

Input known values

Substitute the known values into the formula. \( I = 5.0 \mathrm{~A} \), \( B = 4.0 \mu \mathrm{T} = 4.0 \times 10^{-6} \mathrm{~T} \), and \( \mu_0 = 4\pi \times 10^{-7} \mathrm{~T} \cdot \mathrm{m/A} \).
04

Calculate the distance \( r \)

Place the values into the rearranged formula: \( r = \frac{(4\pi \times 10^{-7} \mathrm{~T} \cdot \mathrm{m/A})(5.0 \mathrm{~A})}{2\pi (4.0 \times 10^{-6} \mathrm{~T})} \).
05

Simplify and find the result

Simplify the equation: \( r = \frac{20\pi \times 10^{-7}}{8\pi \times 10^{-6}} = \frac{20}{8} \times 10^{-1} \mathrm{~m} \), which simplifies further to \( 2.5 \times 10^{-1} \mathrm{~m} \), or 0.25 meters.

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Key Concepts

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

Magnetic Field Calculation
Calculating the magnetic field can seem complex, but it's really all about understanding a simple formula. For a long straight wire carrying a current, the magnetic field around it is given by the formula: \[ B = \frac{\mu_0 I}{2\pi r} \]Here,
  • \( B \) is the magnetic field strength you're trying to calculate.
  • \( \mu_0 \) is a constant called the permeability of free space.
  • \( I \) is the current flowing through the wire.
  • \( r \) is the distance you want to find from the wire where the field is measured.
Once you know the value of the magnetic field, the distance can be easily calculated by rearranging the formula to find \( r \): \[ r = \frac{\mu_0 I}{2\pi B} \] This equation shows how the strength of the current and the distance from the wire influence the magnetic field experienced.
Current-Carrying Wire
A current-carrying wire creates a magnetic field around it. Picture a long wire with a flow of electric charge through it. This movement generates a circular magnetic field that envelops the wire, similar to ripples in a pond. The direction of the magnetic field can be determined using the right-hand rule. Curl your fingers in the direction of the current flow, and your thumb will point in the direction the magnetic field circles the wire.As you can imagine, the strength of the magnetic field is affected by the current. Greater current equals a stronger magnetic field. Additionally, the further you move from the wire, the weaker the field becomes. This relationship is key and is why the distance calculation involves a division by \( r \) in the magnetic field formula.
Physics Lab Experiment
In physics labs, experiments involving magnetic fields are common and enlightening. Students may conduct experiments similar to the one described in the original exercise to grasp fundamental concepts. They usually involve placing sensors at various distances from a wire carrying a specific current and measuring the magnetic field strength. This hands-on experience helps solidify how theoretical equations apply in the real world. By calculating the magnetic field strength and understanding its relationship with distance and current, students can see physics in action and ensure their comprehension of these principles is robust.
Permeability of Free Space
The permeability of free space, denoted as \( \mu_0 \), plays a critical role in magnetic field equations. It is a constant that relates magnetic field strength to the physical properties of space, specifically in a vacuum. The value of \( \mu_0 \) is \( 4\pi \times 10^{-7} \mathrm{~T} \cdot \mathrm{m/A} \). This constant allows us to calculate how magnetic fields behave in empty space and helps bridge the gap between theory and practice.Without \( \mu_0 \), magnetic field calculations would not be consistent, as it sets a standard reference for the vacuum's ability to support magnetic fields. Understanding this constant ensures you can effectively use formulas that involve magnetic fields, as we saw in the original physics exercise to calculate the distance from a current-carrying wire.

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Most popular questions from this chapter

Two long, parallel wires separated by \(50 \mathrm{~cm}\) each carry currents of \(4.0 \mathrm{~A}\) in a horizontal direction. Find the magnetic field midway between the wires if the currents are (a) in the same direction and (b) in opposite directions.

(a) What angle(s) does a particle's velocity have to make with the magnetic field direction for the particle to be subjected to half the maximum possible magnetic force, \(F_{\max } ?\) (b) Express the magnetic force on a charged particle in terms of \(F_{\max }\) if the angle between its velocity and the magnetic field direction is (i) \(10^{\circ},\) (ii) \(80^{\circ},\) and (iii) \(100^{\circ} .\) (c) If the particle's velocity makes an angle of \(50^{\circ}\) with respect to the magnetic field direction, at what other angle(s) would the magnetic force on it be the same? Would the direction be the same? Explain.

A particle with charge \(q\) and mass \(m\) moves in a horizontal plane at right angles to a uniform vertical magnetic field \(B\). (a) What are the period \(T\) and frequency \(f\) of the particle's circular motion in terms of \(q, B\), and \(m\) ? (This frequency is called the cyclotron frequency.) You result should verify that the time for one orbit for any charged particle in a uniform magnetic field is independent of its speed and radius. (b) Compute the path radius and the cyclotron frequency if the particle is an electron with a speed of \(1.0 \times 10^{5} \mathrm{~m} / \mathrm{s}\) traveling in a region where the field strength is \(0.10 \mathrm{mT}\)

Two long, straight, parallel wires \(10 \mathrm{~cm}\) apart carry currents in opposite directions. (a) Use the right-hand source and force rules to determine whether the forces on the wires are (1) attractive or (2) repulsive. Show your reasoning. (b) If the wires carry equal currents of \(3.0 \mathrm{~A}\), what is the magnetic field magnitude that each produces at the other's location? (c) Use the result of part (b) to determine the magnitude of the force per unit length they exert on each other.

A cylindrical solenoid \(10 \mathrm{~cm}\) long has 3000 turns of wire and carries a current of 5.0 A. A second solenoid, consisting of 2000 turns of wire and the same length as the first solenoid, surrounds it and is concentric (shares a common central axis) with it. The outer coil carries a current of 10 A in the same direction as the current in inner one. (a) Find the magnetic field near their common center. (b) What current in the second solenoid (magnitude and relative direction) would make the net field strength at the center twice that of the first solenoid alone? (c) What current in the second solenoid (magnitude and relative direction) would result in zero net magnetic field near their common center?

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