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Chapter 28: Problem 68

Two long, straight parallel wires are separated by a distance of \(20.0 \mathrm{~cm}\). Each wire carries a current of \(10.0 \mathrm{~A}\) in the same direction. What is the magnitude of the resulting magnetic field at a point that is \(12.0 \mathrm{~cm}\) from each wire?

Short Answer

Expert verified
Answer: The total magnetic field at the point is 3.34 × 10⁻⁴ T.

Step by step solution

01

Identify the given values

In this exercise, we are given the current \(I\) carried by both wires as \(10.0 \mathrm{~A}\), the distance between the wires \(d = 20.0 \mathrm{~cm}\), and the distance from each wire to the point where the magnetic field is being measured, \(r = 12.0 \mathrm{~cm}\).
02

Convert the distances to meters

To use the formula for the magnetic field, we need to convert the given distances from centimeters to meters: \(d = 0.20 \mathrm{~m}\) and \(r = 0.12 \mathrm{~m}\).
03

Calculate the magnetic field created by each wire

Using the formula for the magnetic field created by a long, straight current-carrying wire, we can calculate the magnetic field created by each wire at the given point: $$B = \frac{\mu_0I}{2\pi r} = \frac{4\pi × 10^{-7} Tm/A × 10.0 A}{2\pi × 0.12 m} = \frac{4\pi × 10^{-6} T}{0.24\pi}$$ Canceling the \(\pi\) gives: $$B = \frac{10^{-5} T}{0.06}$$ Now calculating the result: $$B = 1.67 × 10^{-4} \mathrm{T}$$ Thus, each wire creates a magnetic field of \(1.67 × 10^{-4} \mathrm{T}\) at the given point.
04

Determine the vector direction of the magnetic fields

Since both wires carry current in the same direction and are parallel to each other, the magnetic fields produced by each wire at the given point will also be in the same direction. This means that the total magnetic field at the point can be found by simply adding the magnitudes of the magnetic fields created by each wire.
05

Calculate the total magnetic field at the given point

Now that we know the magnetic field created by each wire at the given point, we can find the total magnetic field by adding the magnitudes of the magnetic fields: $$B_\text{total} = B_1 + B_2 = 1.67 \times 10^{-4} T + 1.67 \times 10^{-4} T = 3.34 \times 10^{-4} \mathrm{T}$$ The total magnetic field at the point that is \(12.0 \mathrm{~cm}\) from each wire is \(\mathbf{3.34 × 10^{-4} T}\).

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Most popular questions from this chapter

A current element produces a magnetic field in the region surrounding it. At any point in space, the magnetic field produced by this current element points in a direction that is a) radial from the current element to the point in space. b) parallel to the current element. c) perpendicular to the current element and to the radial direction.

Can an ideal solenoid, one with no magnetic field outside the solenoid, exist? If not, does that render the derivation of the magnetic field inside the solenoid (Section 28.4) void?

If you want to construct an electromagnet by running a current of 3.0 A through a solenoid with 500 windings and length \(3.5 \mathrm{~cm}\) and you want the magnetic field inside the solenoid to have magnitude \(B=2.96 \mathrm{~T}\), you can insert a ferrite core into the solenoid. What value of the relative magnetic permeability should this ferrite core have in order to make this work?

A long, straight wire carrying a 2.00-A current lies along the \(x\) -axis. A particle with charge \(q=-3.00 \mu \mathrm{C}\) passes parallel to the \(z\) -axis through the point \((x, y, z)=(0,2,0)\). Where in the \(x y\) -plane should another long, straight wire be placed so that there is no magnetic force on the particle at the point where it crosses the plane?

In a coaxial cable, the solid core carries a current \(i\) The sheath also carries a current \(i\) but in the opposite direction and has an inner radius \(a\) and an outer radius b. The current density is equally distributed over each conductor. Find an expression for the magnetic field at a distance \(a
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