Problem 84 - College Physics

Question

A long, straight horizontal wire carries a current of 2.50 \(\mathrm{A}\) directed toward the right. An electron is traveling in the vicinity of this wire. (a) At the instant the electron is 4.50 \(\mathrm{cm}\) above the wire's center and moving with a speed of \(6.00 \times 10^{4} \mathrm{m} / \mathrm{s}\) directly toward it, what are the magnitude and direction of the force that the magnetic field of the current exerts on the electron? (b) What would be the magnitude and direction of the magnetic force if the electron were instead moving parallel to the wire in the same direction as the current?

Step-by-Step Solution

Answer

(a) 1.07 × 10^{-19} N downward; (b) 0 N.

Step 1: Understanding the Problem

We have a straight wire with current, and an electron moving either towards the wire or parallel to it. We need to find the magnetic force on the electron in both scenarios using the magnetic field generated by the wire.

Step 2: Formula for Magnetic Field by a Straight Wire

The magnetic field (B) at a distance r from a long straight wire with current I is given by: \[ B = \frac{\mu_0 I}{2 \pi r} \]where \( \mu_0 = 4\pi \times 10^{-7} \ \mathrm{T\cdot m/A} \) is the permeability of free space.

Step 3: Calculate Magnetic Field at Electron's Position in Scenario (a)

Here, \( I = 2.50 \, \mathrm{A} \) and \( r = 4.50 \, \mathrm{cm} = 0.045 \, \mathrm{m} \). Calculate B:\[ B = \frac{(4\pi \times 10^{-7})(2.50)}{2\pi(0.045)} \approx 1.11 \times 10^{-5} \, \mathrm{T} \]

Step 4: Determine the Force on the Electron in Scenario (a)

Use the Lorentz force formula: \( F = qvB \sin\theta \), where \( q = -1.6 \times 10^{-19} \, \mathrm{C} \) for an electron, \( v = 6.00 \times 10^4 \, \mathrm{m/s} \), and since the direction is perpendicular (\( \theta = 90^\circ \), \( \sin 90^\circ = 1 \)).\[ F = (-1.6 \times 10^{-19})(6.00 \times 10^4)(1.11 \times 10^{-5})(1) \approx -1.07 \times 10^{-19} \mathrm{N} \]

Step 5: Direction of Force in Scenario (a)

Using the right-hand rule, point your thumb in the direction of the electron's velocity (toward the wire) and fingers in the direction of the magnetic field (into the page). Your palm points out, but since the electron is negative, the force direction is opposite (out of the page). Thus, force is downward.

Step 6: Electron's Motion Parallel to Current in Scenario (b)

When the electron moves parallel to the current, \( \theta = 0^\circ \), so \( \sin\theta = 0 \). Therefore, \[ F = qvB \sin 0 \] is zero because the velocity direction and magnetic field are parallel.

Step 7: Conclusion

In Scenario (a), the force on the electron is approximately \( 1.07 \times 10^{-19} \, \mathrm{N} \) downward. In Scenario (b), the force is 0.

Key Concepts

Magnetic Field

A magnetic field is a region in space where a magnetic force is exerted on charged particles. A current-carrying wire, like the one in our problem, generates a magnetic field around it. The strength of this magnetic field at a distance from a long straight wire can be calculated using the formula:

• \[ B = \frac{\mu_0 I}{2 \pi r} \]
• Where \( B \) is the magnetic field, \( I \) is the current in the wire, \( r \) is the distance from the wire, and \( \mu_0 \) is the permeability of free space, approximately \( 4\pi \times 10^{-7} \ \mathrm{T\cdot m/A} \).

The field's direction is determined using the right-hand rule, which we'll explore further below. But it's crucial to understand that the magnetic field influences how charged particles like electrons move when they're near a wire.

Lorentz Force

The Lorentz force is the force experienced by a charged particle moving in a magnetic field. When an electron or any charged particle with charge \( q \) and velocity \( v \) moves in a magnetic field \( B \), it experiences a force given by the equation:

• \[ F = qvB \sin \theta \]
• Here, \( \theta \) is the angle between the velocity of the charge and the magnetic field direction.

In scenario (a) of our exercise, the electron is moving toward the wire at a right angle to the magnetic field, meaning \( \theta = 90^\circ \). As \( \sin 90^\circ = 1 \), the equation simplifies to \( F = qvB \). The calculated force would point to a specific direction, influenced by the charge of the electron and its interaction with the magnetic field.

Right-Hand Rule

The right-hand rule is a tool to determine the direction of the magnetic force, current, or magnetic field. Let's focus on calculating the direction of the force where the electron is the test charge.

• Point your thumb in the direction of the charge's velocity.
• Your fingers should curl in the direction of the magnetic field lines.
• Your palm then faces the direction of the force on a positive charge (opposite for negative charge like an electron).

In the exercise, when the electron moves toward the wire with the magnetic field directed into the page, your palm faces outwards. Because the electron has a negative charge, reverse this direction to determine the force is actually downward.

Current-Carrying Wire

A wire carrying an electric current creates a magnetic field around it. In the exercise, the wire has a current of \( 2.50 \, \mathrm{A} \) directed toward the right. The generated magnetic field affects any charged particles nearby, such as the electron in the problem. When calculating the magnetic field around a long, straight, current-carrying wire using the formula previously mentioned, you can predict the behavior of any charged particle within its vicinity. This is because the field circles the wire in closed loops dictated by the right-hand rule, impacting electron paths and interacting with their charges to exert force.

Electron Motion

Electron motion in a magnetic field is subject to changes in direction due to the Lorentz force. The electron will continue to experience this force as long as it remains within the magnetic field's influence.

• In scenario (a), the electron heads toward the wire and is deflected downward due to the magnetic force exerted by the field from the wire.
• In scenario (b), when the electron's motion is parallel to the wire, no magnetic force acts on it. This is because the velocity and magnetic field direction are parallel, leading to \( \theta = 0^\circ \) and subsequently \( \sin \theta = 0 \).

Therefore, the force in this case is zero and the electron continues in a straight path, unaffected by the field.