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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 charged particle is moving in a constant magnetic field. State whether each of the following statements concerning the magnetic force exerted on the particle is true or false? (Assume that the magnetic field is not parallel or antiparallel to the velocity.) a) It does no work on the particle. b) It may increase the speed of the particle. c) It may change the velocity of the particle. d) It can act only on the particle while the particle is in motion. e) It does not change the kinetic energy of the particle.

A circular coil with a radius of \(10.0 \mathrm{~cm}\) has 100 turns of wire and carries a current, \(i=100 . \mathrm{mA} .\) It is free to rotate in a region with a constant horizontal magnetic field given by \(\vec{B}=(0.0100 \mathrm{~T}) \hat{x}\). If the unit normal vector to the plane of the coil makes an angle of \(30.0^{\circ}\) with the horizontal, what is the magnitude of the net magnetic torque acting on the coil? 27.61 At \(t=0\) an electron crosses the positive \(y\) -axis (so \(x=0\) ) at \(60.0 \mathrm{~cm}\) from the origin with velocity \(2.00 \cdot 10^{5} \mathrm{~m} / \mathrm{s}\) in the positive \(x\) -direction. It is in a uniform magnetic field. a) Find the magnitude and the direction of the magnetic field that will cause the electron to cross the \(x\) -axis at \(x=60.0 \mathrm{~cm}\). b) What work is done on the electron during this motion? c) How long will the trip take from \(y\) -axis to \(x\) -axis?

A coil consists of 120 circular loops of wire of radius \(4.8 \mathrm{~cm} .\) A current of 0.49 A runs through the coil, which is oriented vertically and is free to rotate about a vertical axis (parallel to the \(z\) -axis). It experiences a uniform horizontal magnetic field in the positive \(x\) -direction. When the coil is oriented parallel to the \(x\) -axis, a force of \(1.2 \mathrm{~N}\) applied to the edge of the coil in the positive \(y\) -direction can keep it from rotating. Calculate the strength of the magnetic field.

An electron is moving with a constant velocity. When it enters an electric field that is perpendicular to its velocity, the electron will follow a ________ trajectory. When the electron enters a magnetic field that is perpendicular to its velocity, it will follow a ________ trajectory.

A square loop of wire of side length \(\ell\) lies in the \(x y\) -plane, with its center at the origin and its sides parallel to the \(x\) - and \(y\) -axes. It carries a current, \(i\), in the counterclockwise direction, as viewed looking down the \(z\) -axis from the positive direction. The loop is in a magnetic field given by \(\vec{B}=\left(B_{0} / a\right)(z \hat{x}+x \hat{z}),\) where \(B_{0}\) is a constant field strength, \(a\) is a constant with the dimension of length, and \(\hat{x}\) and \(\hat{z}\) are unit vectors in the positive \(x\) -direction and positive \(z\) -direction. Calculate the net force on the loop.
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