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Chapter 19: Problem 21

An electron moves with speed \(2.0 \times 10^{5} \mathrm{m} / \mathrm{s}\) in a uniform magnetic field of \(1.4 \mathrm{T},\) pointing south. At one instant, the electron experiences an upward magnetic force of $1.6 \times 10^{-14} \mathrm{N} .$ In what direction is the electron moving at that instant? 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 possibilitics.)

Short Answer

Expert verified
Answer: The electron is moving towards the West.

Step by step solution

01

Find the charge of an electron

Recall that the charge of an electron is: \(q = -1.6 \times 10^{-19} \,\text{C}\)
02

Write the formula for magnetic force

Write the formula for the magnitude of the magnetic force acting on a charged particle: \(|\vec{F}| = q|\vec{v}| |\vec{B}| \sin\theta\)
03

Plug in the given values

Replace the magnitudes of the force, charge, velocity, and the magnetic field with the given values: \(1.6 \times 10^{-14} \,\text{N} = (-1.6 \times 10^{-19} \,\text{C} ) (2.0 \times 10^{5} \frac{\text{m}}{\text{s}}) (1.4 \, \text{T}) \sin\theta\)
04

Solve for theta

Rearrange and solve for \(\theta\): \(\sin\theta = \frac{1.6 \times 10^{-14} \,\text{N}}{ (-1.6 \times 10^{-19} \,\text{C} ) (2.0 \times 10^{5} \frac{\text{m}}{\text{s}}) (1.4 \, \text{T})}\) Calculate the value of \(\sin\theta\): \(\sin\theta = \frac{1.6 \times 10^{-14}}{(-1.6 \times 10^{-19})(2.0 \times 10^5)(1.4)} = -1\) Now find the angle \(\theta\): \(\theta = \arcsin (-1)\)
05

Determine all possible angles

For a sine function, \(-1\) occurs at \(270^{\circ}\) or \(7 \pi / 2\). Therefore, \(\theta = 270^{\circ}\) or \(7 \pi / 2\). Since \(\theta\) is \(270^{\circ}\), it means the direction of the electron's velocity is \(270^{\circ}\) from south, which is towards west. So, the electron is moving towards the West at that instant.

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Most popular questions from this chapter

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