An electron of mass \(m\) and charge \(-e\) is traveling to the right parallel to a wire with speed \(v\). The electron is a distance \(d\) from the wire. The wire is carrying a current \(I\) to the right, as shown in the figure above.
Which of the following gives the magnitude and direction of the force exerted on the electron by the current-carrying wire?
| Magnitude | Direction | |
|---|---|---|
| (A) | \( \dfrac{\mu_{0}Iev}{2\pi d} \) | Toward the top of the page |
| (B) | \( \dfrac{\mu_{0}Iev}{2\pi d} \) | Out of the page |
| (C) | \( \dfrac{\mu_{0}Iev}{2\pi d} \) | Into the page |
| (D) | \( \dfrac{\mu_{0}Iev}{2m\pi d} \) | Toward the top of the page |
| (E) | \( \dfrac{\mu_{0}Iev}{2m\pi d} \) | Out of the page |
Correct Answer: \( \boxed{\mathrm{A}} \)
The magnetic field produced by a long straight current-carrying wire is
\( B=\dfrac{\mu_{0}I}{2\pi d} \).
The magnetic force on the moving electron has magnitude
\( F=|q|vB \),
since the electron’s velocity is perpendicular to the magnetic field at its location.
Substituting the expression for \(B\),
\( F=e v\left(\dfrac{\mu_{0}I}{2\pi d}\right)=\dfrac{\mu_{0}Iev}{2\pi d}. \)
Using the right-hand rule, the magnetic field produced by the wire at a point above the wire is directed out of the page.
For a positive charge moving to the right, \( \vec{v}\times\vec{B} \) points toward the bottom of the page. Since the particle is an electron, the force is in the opposite direction, namely toward the top of the page.
Therefore, the correct answer is (A).
A narrow beam of protons produces a current of \(1.6\times10^{-3}\,\mathrm{A}\). There are \(10^{9}\) protons in each meter along the beam.
Which of the following describes the lines of magnetic field in the vicinity of the beam due to the beam’s current?
Correct Answer: \( \boxed{\mathrm{A}} \)
A moving beam of protons constitutes an electric current. A straight current-carrying conductor produces magnetic field lines that form closed concentric circles around the direction of the current.
The direction of the magnetic field is determined using the right-hand rule: point the thumb of the right hand in the direction of the conventional current (the proton beam), and the curled fingers indicate the direction of the magnetic field.
The magnitude of the magnetic field at a distance \(r\) from a long straight current is
\( B=\dfrac{\mu_0 I}{2\pi r} \)
Therefore, the magnetic field lines are concentric circles surrounding the beam.
Hence, the correct answer is (A).
Two parallel wires, A and B, have currents in opposite directions, as shown in the figure above. Current \(i_{B}\) is twice as large as \(i_{A}\). The force on wire A due to current \(i_{B}\) has magnitude \(F\).
Which of the following correctly describes the direction and magnitude of the force on wire B due to current \(i_{A}\)?
| Direction | Magnitude | |
|---|---|---|
| (A) | To the left | \(F\) |
| (B) | To the left | \(2F\) |
| (C) | To the left | \(4F\) |
| (D) | To the right | \(F\) |
| (E) | To the right | \(2F\) |
Correct Answer: \( \boxed{\mathrm{D}} \)
The magnetic force per unit length between two long parallel current-carrying wires is
\( \dfrac{F}{L}=\dfrac{\mu_{0}i_{A}i_{B}}{2\pi d}, \)
where \(d\) is the separation between the wires.
Since the currents flow in opposite directions, the wires repel each other.
Therefore, wire A is pushed to the left, and wire B is pushed to the right.
By Newton’s Third Law, the force that wire A exerts on wire B has the same magnitude as the force that wire B exerts on wire A, even though the currents are different.
Thus,
\( |\vec{F}_{BA}|=|\vec{F}_{AB}|=F. \)
Hence, wire B experiences a force of magnitude \(F\) directed to the right.
Therefore, the correct answer is (D).
