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Chapter 19: Problem 34

Crossed electric and magnetic fields are established over a certain region. The magnetic field is 0.635 T vertically downward. The electric field is $2.68 \times 10^{6} \mathrm{V} / \mathrm{m}$ horizontally east. An electron, traveling horizontally northward, experiences zero net force from these fields and so continues moving in a straight line. What is the electron's speed?

Short Answer

Expert verified
Answer: The speed of the electron is 4.22 x 10^6 m/s.

Step by step solution

01

Write down the given values

The given values are: Magnetic field strength, \(B = 0.635\, T\) (vertical, downward) Electric field strength, \(E = 2.68 \times 10^6\, \mathrm{V/m}\) (horizontal, east) Electron charge, \(q = -1.6 \times 10^{-19}\) C Desired force on the electron, \(F_{net} = 0\)
02

Equate the electric and magnetic forces

The net force on the electron is zero. Therefore, the electric force and the magnetic force should be equal and opposite: \(F_E = F_B\) Since \(F_E=qE\) and \(F_B=qvB\sin\theta\), we can write: \(qE = qvB\sin\theta\)
03

Determine the angle between the velocity and the magnetic field

In this problem, the electron is traveling horizontally northward, and the magnetic field is vertically downward. The angle between these two directions is 90 degrees, i.e., \(\theta = 90^{\circ}\) or \(\sin\theta = 1\).
04

Solve for the electron's speed

Now we can solve for the electron's speed, v: \(E = vB\sin\theta\) \(2.68 \times 10^6\, \mathrm{V/m} = v (0.635\, T)(1)\) \(v = \frac{2.68 \times 10^6\, \mathrm{V/m}}{0.635\, T}\) \(v \approx 4.22 \times 10^6\, \mathrm{m/s}\) Therefore, the speed of the electron is \(\boxed{4.22 \times 10^6\, \mathrm{m/s}}\).

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Most popular questions from this chapter

A straight wire segment of length \(0.60 \mathrm{m}\) carries a current of $18.0 \mathrm{A}$ and is immersed in a uniform external magnetic ficld of magnitude \(0.20 \mathrm{T}\). (a) What is the magnitude of the maximum possible magnetic force on the wire segment? (b) Explain why the given information enables you to calculate only the maximum possible force.
A strip of copper carries current in the \(+x\) -direction. There is an external magnctic field directed out of the page. What is the direction of the Hall electric field?
The torque on a loop of wire (a magnetic dipole) in a uniform magnetic field is \(\tau=N I A B\) sin \(\theta,\) where \(\theta\) is the angle between \(\overline{\mathbf{B}}\) and a line perpendicular to the loop of wire. Suppose an electric dipole, consisting of two charges \(\pm q\) a fixed distance \(d\) apart is in a uniform electric field \(\overrightarrow{\mathbf{E}}\). (a) Show that the net electric force on the dipole is zero. (b) Let \(\theta\) be the angle between \(\overrightarrow{\mathbf{E}}\) and a line running from the negative to the positive charge. Show that the torque on the electric dipole is \(\tau=q d E\) sin \(\theta\) for all angles $-180^{\circ} \leq \theta \leq 180^{\circ} .$ (Thus, for both electric and magnetic dipoles, the torque is the product of the dipole moment times the field strength times $\sin \theta .\( The quantity \)q d\( is the electric dipole moment; the quantity \)N I A$ is the magnetic dipole moment.)
A strip of copper \(2.0 \mathrm{cm}\) wide carries a current \(I=\) $30.0 \mathrm{A}\( to the right. The strip is in a magnetic field \)B=5.0 \mathrm{T}$ into the page. (a) What is the direction of the average magnetic force on the conduction electrons? (b) The Hall voltage is \(20.0 \mu \mathrm{V}\). What is the drift velocity?
Two identical bar magnets lie next to one another on a table. Sketch the magnetic field lines if the north poles are at opposite ends.
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