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Chapter 19: Problem 8

Find the magnetic force exerted on an electron moving vertically upward at a speed of \(2.0 \times 10^{7} \mathrm{m} / \mathrm{s}\) by a horizontal magnetic field of \(0.50 \mathrm{T}\) directed north.

Short Answer

Expert verified
Question: Calculate the magnetic force exerted on an electron moving vertically upward with a speed of \(2.0 \times 10^{7} \mathrm{m} / \mathrm{s}\) through a \(0.50 \mathrm{T}\) magnetic field directed north. Answer: The magnetic force on the electron is approximately \(-1.6 \times 10^{-12} \mathrm{N}\).

Step by step solution

01

Identify known values.

The values that are provided in the question are: 1. Speed of electron (v) : \(2.0 \times 10^{7} \mathrm{m} / \mathrm{s}\) (vertically upward) 2. Magnetic field (B) : \(0.50 \mathrm{T}\) (directed north) 3. Electron charge (q) : \(-1.6 \times 10^{-19} \mathrm{C}\) (since it's an electron) Since the electron is moving vertically upward and the magnetic field is directed north (horizontal), the angle between the velocity vector and the magnetic field vector (θ) is 90 degrees.
02

Calculate the magnetic force using the formula F = qvBsinθ.

Now, we can use the formula for magnetic force on a moving charge: F = qvBsinθ Here, we already have all the known values: q = \(-1.6 \times 10^{-19} \mathrm{C}\) v = \(2.0 \times 10^{7} \mathrm{m} / \mathrm{s}\) B = \(0.50 \mathrm{T}\) θ = 90° (and sin90° = 1) Plugging in the known values into the formula, we get: F = \((-1.6 \times 10^{-19} \mathrm{C}) (2.0 \times 10^{7} \mathrm{m} / \mathrm{s}) (0.50 \mathrm{T}) (1)\)
03

Calculate the magnetic force F.

Performing the multiplication will give us the magnetic force acting on the electron: F ≈ \(-1.6 \times 10^{-19} \mathrm{C} \times 2.0 \times 10^{7} \mathrm{m} / \mathrm{s} \times 0.50 \mathrm{T}\) F ≈ \(-1.6 \times 10^{-19} \mathrm{C} \times 1.0 \times 10^{7} \mathrm{m} / \mathrm{s} \times \mathrm{T}\) F ≈ \(-1.6 \times 10^{-12} \mathrm{N}\) The magnetic force on the electron is approximately \(-1.6 \times 10^{-12} \mathrm{N}\). The negative sign indicates that the force is in the opposite direction to the reference direction.

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Most popular questions from this chapter

A singly charged ion of unknown mass moves in a circle of radius $12.5 \mathrm{cm}\( in a magnetic field of \)1.2 \mathrm{T}$. The ion was accelerated through a potential difference of \(7.0 \mathrm{kV}\) before it entered the magnetic field. What is the mass of the ion?
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?
A straight wire is aligned east-west in a region where Earth's magnetic field has magnitude \(0.48 \mathrm{mT}\) and direction \(72^{\circ}\) below the horizontal, with the horizontal component directed due north. The wire carries a current \(I\) toward the west. The magnetic force on the wire per unit length of wire has magnitude \(0.020 \mathrm{N} / \mathrm{m}\) (a) What is the direction of the magnetic force on the wire? (b) What is the current \(I ?\)
The intrinsic magnetic dipole moment of the electron has magnitude $9.3 \times 10^{-24} \mathrm{A} \cdot \mathrm{m}^{2} .$ In other words, the electron acts as though it were a tiny current loop with $N I A=9.3 \times 10^{-24} \mathrm{A} \cdot \mathrm{m}^{2} .$ What is the maximum torque on an electron due to its intrinsic dipole moment in a \(1.0-T\) magnetic field?
The conversion between atomic mass units and kilograms is $$1 \mathrm{u}=1.66 \times 10^{-27} \mathrm{kg}$$ A sample containing carbon (atomic mass 12 u), oxygen \((16 \mathrm{u}),\) and an unknown element is placed in a mass spectrometer. The ions all have the same charge and are accelerated through the same potential difference before entering the magnetic field. The carbon and oxygen lines are separated by \(2.250 \mathrm{cm}\) on the photographic plate, and the unknown element makes a line between them that is \(1.160 \mathrm{cm}\) from the carbon line. (a) What is the mass of the unknown clement? (b) Identify the element.
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