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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

An electron moves with speed \(2.0 \times 10^{5} \mathrm{m} / \mathrm{s}\) in a \(1.2-\mathrm{T}\) uniform magnctic ficld. At one instant, the electron is moving due west and experiences an upward magnetic force of $3.2 \times 10^{-14} \mathrm{N} .$ What is the direction of the magnetic field? Be specific: give the angle(s) with respect to $\mathrm{N}, \mathrm{S}, \mathrm{E}, \mathrm{W},$ up, down. (If there is more than one possible answer, find all the possibilities.)
The magnetic field in a cyclotron is \(0.360 \mathrm{T}\). The dees have radius \(82.0 \mathrm{cm} .\) What maximum speed can a proton achicve in this cyclotron?
In a certain region of space, there is a uniform electric field \(\overrightarrow{\mathbf{E}}=3.0 \times 10^{4} \mathrm{V} / \mathrm{m}\) directed due east and a uniform magnetic field \(\mathbf{B}=0.080\) T also directed due east. What is the electromagnetic force on an electron moving due south at \(5.0 \times 10^{6} \mathrm{m} / \mathrm{s} ?\)
The conversion between atomic mass units and kilograms is $$1 \mathrm{u}=1.66 \times 10^{-27} \mathrm{kg}$$ In one type of mass spectrometer, ions having the same velocity move through a uniform magnetic field. The spectrometer is being used to distinguish $^{12} \mathrm{C}^{+}\( and \)^{14} \mathrm{C}^{+}$ ions that have the same charge. The \(^{12} \mathrm{C}^{+}\) ions move in a circle of diameter \(25 \mathrm{cm} .\) (a) What is the diameter of the orbit of \(^{14} \mathrm{C}^{+}\) ions? (b) What is the ratio of the frequencies of revolution for the two types of ion?
At a certain point on Earth's surface in the southern hemisphere, the magnetic field has a magnitude of \(5.0 \times 10^{-5} \mathrm{T}\) and points upward and toward the north at an angle of \(55^{\circ}\) above the horizontal. A cosmic ray muon with the same charge as an electron and a mass of $1.9 \times 10^{-28} \mathrm{kg}$ is moving directly down toward Earth's surface with a speed of \(4.5 \times 10^{7} \mathrm{m} / \mathrm{s} .\) What is the magnitude and direction of the force on the muon?
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