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Chapter 15: Problem 37

Two very long, straight, parallel wires carry steady current \(I\) and \(-I\), respectively. The distance between the wires is \(d\). At a certain instant of time, a point charge \(q\) is at a point equidistant from the two wires and in the plane of the wires. Its instantaneous velocity \(\vec{v}\) is perpendicular to this plane. The magnitude of the force due to the magnetic field acting on the charge at this instant is (A) \(\frac{\mu_{0} I q v}{2 \pi d}\) (B) \(\frac{\mu_{0} I q v}{\pi d}\) (C) \(\frac{2 \mu_{0} I q v}{\pi d}\) (D) Zero

Short Answer

Expert verified
The short answer based on the step-by-step solution is: The magnitude of the force due to the magnetic field acting on the charge at that instant is \(F = \frac{2 \mu_{0} I q v}{\pi d}\). The correct answer is (C) \(\frac{2 \mu_{0} I q v}{\pi d}\).

Step by step solution

01

Find the magnetic field due to each wire

First, we find the magnetic field due to each wire. The magnetic field at a distance \(r\) from a long, straight wire carrying a current \(I\) is given by the Biot-Savart Law: \[B = \frac{\mu_{0}I}{2\pi r}\] The magnetic field due to the first wire carrying current \(I\) is: \[B_{1} = \frac{\mu_{0}I}{2\pi r_{1}}\] And the magnetic field due to the second wire carrying current \(-I\) is: \[B_{2} = \frac{\mu_{0}(-I)}{2\pi r_{2}}\]
02

Determine the distance from the charge to each wire

The point charge \(q\) is equidistant from the two wires, so the distances, \(r_{1}\) and \(r_{2}\), are both equal to \(\frac{d}{2}\).
03

Calculate the net magnetic field at the location of the charge

Since the point charge is equidistant from both wires and in the plane of the wires, the magnetic fields due to each wire will be in opposite directions. Thus, we can find the net magnetic field by taking the difference between the magnitudes of the individual magnetic fields: \[B_{net} = |B_{1} - B_{2}| = \left|\frac{\mu_{0}I}{2\pi (\frac{d}{2})} - \frac{\mu_{0}(-I)}{2\pi (\frac{d}{2})} \right|\]
04

Simplify the expression for the net magnetic field

Simplifying the expression for the net magnetic field, we get: \[B_{net} = \frac{\mu_{0}I}{\pi d} + \frac{\mu_{0}I}{\pi d} = \frac{2 \mu_{0}I}{\pi d}\]
05

Calculate the magnetic force acting on the charge

Now, we can use the formula for the magnetic force on a moving charge to calculate the magnitude of the force: \[F = qvB_{net}\sin{\theta}\] Since the velocity vector, \(\vec{v}\), of the charge is perpendicular to the plane of the wires (and thus perpendicular to the magnetic field), \(\theta = 90^{\circ}\). Therefore, \(\sin{\theta} = 1\), and the formula simplifies to: \[F = qvB_{net}\]
06

Substitute the expression for the net magnetic field and find the force

Substitute the expression for \(B_{net}\): \[F = qv\frac{2 \mu_{0}I}{\pi d}\] So the magnitude of the force due to the magnetic field acting on the charge at that instant is \[F = \frac{2 \mu_{0} I q v}{\pi d}\] Thus, the correct answer is (C) \(\frac{2 \mu_{0} I q v}{\pi d}\).

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Most popular questions from this chapter

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Two long parallel wires \(P\) and \(Q\) are held perpendicular to the plane of the paper at a separation of \(5 \mathrm{~m}\). If \(P\) and \(Q\) carry currents of \(2.5 \mathrm{~A}\) and \(5 \mathrm{~A}\), respectively, in the same direction, then the magnetic field at a point midway between \(P\) and \(Q\) is (A) \(\frac{\mu_{0}}{\pi}\) (B) \(\frac{\sqrt{3} \mu_{0}}{\pi}\) (C) \(\frac{\mu_{0}}{2 \pi}\) (d) \(\frac{3 \mu_{0}}{2 \pi}\)
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