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Chapter 27: Problem 6

In which direction does a magnetic force act on an electron that is moving in the positive \(x\) -direction in a magnetic field pointing in the positive \(z\) -direction? a) the positive \(y\) -direction b) the negative \(y\) -direction c) the negative \(x\) -direction d) any direction in the \(x y\) -plane

Short Answer

Expert verified
Answer: b) the negative y-direction. Explanation: Using the Lorentz force equation and the right-hand rule, the magnetic force acts in the positive y-direction for a positive charge, but since the electron has a negative charge, the force is in the opposite direction, i.e., the negative y-direction.

Step by step solution

01

Determine the electron's charge

An electron has a negative charge, which can be represented as \(q = -e\), where \(e\) is the elementary charge (\(1.6 \times 10^{-19}\) Coulombs).
02

Write down the known quantities/vector components

An electron is moving in the positive \(x\)-direction, so its velocity vector is \(\vec{v} = v_x\hat{i}\), where \(v_x > 0\). The magnetic field is in the positive \(z\)-direction, so the magnetic field vector is \(\vec{B} = B_z\hat{k}\), where \(B_z > 0\).
03

Compute the cross product

The magnetic force acting on the electron is given by the Lorentz force equation: \[ \vec{F} = q(\vec{v} \times \vec{B}) \] Since \(q=-e\), we have: \[ \vec{F} = -e(\vec{v} \times \vec{B}) \] To take the cross product, we can use the determinant formula: \[ \vec{F} = -e\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ v_x & 0 & 0 \\ 0 & 0 & B_z \end{vmatrix} \] Applying the determinant formula, we find: \[ \vec{F} = -e(\hat{i}(0 - 0) - \hat{j}(0 - 0) + \hat{k}(0 - 0)) = -e\cdot 0 \] So, the magnetic force is zero in the \(x\) and \(z\) directions.
04

Find the direction of the magnetic force in the \(y\)-direction

Now we need to use the right-hand rule to find the direction of the magnetic force in the \(y\)-direction. Point your right-hand fingers in the direction of the electron's velocity (\(+x\)), and then curl them toward the direction of the magnetic field (\(+z\)). Your thumb will be pointing in the direction of the force. Since the electron has a negative charge, the actual force will be in the opposite direction. In this case, your thumb will be pointing in the positive \(y\)-direction, but since the electron has a negative charge, the force is in the negative \(y\)-direction. Therefore, the correct answer is: b) the negative \(y\)-direction

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Most popular questions from this chapter

A current-carrying wire is positioned within a large, uniform magnetic field, \(\vec{B}\). However, the wire experiences no force. Explain how this might be possible.

A particle with mass \(m\), charge \(q\), and velocity \(v\) enters a magnetic field of magnitude \(B\) and with direction perpendicular to the initial velocity of the particle. What is the work done by the magnetic field on the particle? How does this affect the particle's motion?

A proton, moving in negative \(y\) -direction in a magnetic field, experiences a force of magnitude \(F\), acting in the negative \(x\) -direction. a) What is the direction of the magnetic field producing this force? b) Does your answer change if the word "proton" in the statement is replaced by “electron"?

An electron is moving at \(v=6.00 \cdot 10^{7} \mathrm{~m} / \mathrm{s}\) perpendicular to the Earth's magnetic field. If the field strength is \(0.500 \cdot 10^{-4} \mathrm{~T}\), what is the radius of the electron's circular path?

A simple galvanometer is made from a coil that consists of \(N\) loops of wire of area \(A .\) The coil is attached to a mass, \(M\), by a light rigid rod of length \(L\). With no current in the coil, the mass hangs straight down, and the coil lies in a horizontal plane. The coil is in a uniform magnetic field of magnitude \(B\) that is oriented horizontally. Calculate the angle from the vertical of the rigid rod as a function of the current, \(i\), in the coil.
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