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Chapter 19: Problem 9

Find the magnetic force exerted on a proton moving east at a speed of $6.0 \times 10^{6} \mathrm{m} / \mathrm{s}\( by a horizontal magnetic ficld of \)2.50 \mathrm{T}$ directed north.

Short Answer

Expert verified
Answer: The magnetic force exerted on the proton is \(2.40 \times 10^{-12} \, N\).

Step by step solution

01

Identify relevant information and formula

We are given the following information: - The velocity of the proton, \(v = 6.0 \times 10^6 \frac{m}{s}\) (eastward) - The magnetic field, \(B = 2.50 \, T\) (northward) - Charge of a proton, \(q = 1.6 \times 10^{-19} \, C\) We will use the formula for the magnetic force on a moving charged particle: \(F = q|\vec{v}||\vec{B}|\sin{\theta}\), where \(\theta\) is the angle between the velocity and magnetic field vectors. In this problem, the proton is moving eastward while the magnetic field is directed northward, which means that the angle between the two is \(\theta = 90^{\circ}\).
02

Calculate the magnetic force

Now plug in the values given into the force equation: \(F = (1.6 \times 10^{-19} \, C)(6.0 \times 10^6 \frac{m}{s})(2.50 \, T)\sin{90^{\circ}}\) Since \(\sin{90^{\circ}} = 1\), the formula simplifies to: \(F = 1.6 \times 10^{-19} \, C \times 6.0 \times 10^6 \frac{m}{s} \times 2.50 \, T\) Multiplying these values gives us the magnetic force exerted on the proton: \(F = 2.40 \times 10^{-12} \, N\)
03

State the final answer

The magnetic force exerted on a proton moving east at a speed of \(6.0 \times 10^6 \frac{m}{s}\) by a horizontal magnetic field of \(2.50 \, T\) directed north is \(2.40 \times 10^{-12} \, N\).

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Most popular questions from this chapter

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 \(I ?\)
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.)
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.
Two long straight wires carry the same amount of current in the directions indicated. The wires cross each other in the plane of the paper. Rank points \(A, B\) \(C,\) and \(D\) in order of decreasing field strength.
In a certain region of space, there is a uniform electric field \(\overrightarrow{\mathbf{E}}=2.0 \times 10^{4} \mathrm{V} / \mathrm{m}\) to the east and a uniform magnetic field \(\overline{\mathbf{B}}=0.0050 \mathrm{T}\) to the west. (a) What is the electromagnetic force on an electron moving north at \(1.0 \times 10^{7} \mathrm{m} / \mathrm{s} ?(\mathrm{b})\) With the electric and magnetic fields as specified, is there some velocity such that the net electromagnetic force on the electron would be zero? If so, give the magnitude and direction of that velocity. If not, explain briefly why not.
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