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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

When a magnetic dipole is placed in a magnetic field, it has a natural tendency to minimize its potential energy by aligning itself with the field. If there is sufficient thermal energy present, however, the dipole may rotate so that it is no longer aligned with the field. Using \(k_{\mathrm{B}} T\) as a measure of the thermal energy, where \(k_{\mathrm{B}}\) is Boltzmann's constant and \(T\) is the temperature in kelvins, determine the temperature at which there is sufficient thermal energy to rotate the magnetic dipole associated with a hydrogen atom from an orientation parallel to an applied magnetic field to one that is antiparallel to the applied field. Assume that the strength of the field is \(0.15 \mathrm{~T}\)

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.

A 50-turn rectangular coil of wire of dimensions \(10.0 \mathrm{~cm}\) by \(20.0 \mathrm{~cm}\) lies in a horizontal plane, as shown in the figure. The axis of rotation of the coil is aligned north and south. It carries a current \(i=1.00 \mathrm{~A}\), and is in a magnetic field pointing from west to east. A mass of \(50.0 \mathrm{~g}\) hangs from one side of the loop. Determine the strength the magnetic field has to have to keep the loop in the horizontal orientation.

The number of turns in a solenoid is doubled, and its length is halved. How does its magnetic field change? a) it doubles b) it is halved c) it quadruples d) it remains unchanged

Many electrical applications use twisted-pair cables in which the ground and signal wires spiral about each other. Why?
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