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Chapter 19: Problem 19

A positron \((q=+e)\) moves at \(5.0 \times 10^{7} \mathrm{m} / \mathrm{s}\) in a magnetic field of magnitude 0.47 T. The magnetic force on the positron has magnitude \(2.3 \times 10^{-12} \mathrm{N} .\) (a) What is the component of the positron's velocity perpendicular to the magnetic field? (b) What is the component of the positron's velocity parallel to the magnetic field? (c) What is the angle between the velocity and the field?

Short Answer

Expert verified
Answer: The component of the positron's velocity perpendicular to the magnetic field is \(7.0\times 10^6 m/s\). The component of the positron's velocity parallel to the magnetic field is \(4.87\times 10^7 m/s\). The angle between the velocity and the magnetic field is \(8.12^\circ\).

Step by step solution

01

Apply the magnetic force formula

We know that the magnitude of magnetic force on a moving charge in a magnetic field is given by the formula \(F = qvB\sin\theta\), where \(F\) is the magnetic force, \(q\) is the charge, \(v\) is the velocity, \(B\) is the magnetic field strength, and \(\theta\) is the angle between the velocity vector and the magnetic field. We are given the force magnitude \(F = 2.3\times 10^{-12}N\), charge \(q = +e\), and magnetic field strength \(B = 0.47T\). We can rearrange the formula to find the perpendicular velocity component, \(v_\perp\).
02

Calculate the perpendicular velocity component

First, we calculate the product \(qvB\), which is equal to \(F / \sin\theta\). Since we know the magnitude of the force \(F\), we can find \(qvB\): \(2.3\times 10^{-12} = e \cdot (5.0\times 10^7) \cdot 0.47 \cdot \sin\theta\) Now we can rearrange the equation to find the perpendicular velocity component: \(v_\perp = \frac{2.3\times 10^{-12}}{e\cdot 0.47} = \frac{2.3\times 10^{-12}}{(1.6\times 10^{-19})\cdot 0.47} = 7.02\times 10^6 m/s\)
03

Calculate the parallel velocity component

Next, we need to find the parallel component of the velocity, \(v_\parallel\). Since we know the total velocity (\(5.0\times 10^7 m/s\)) and the perpendicular component (\(7.0\times 10^6 m/s\)), we can use the Pythagorean theorem to find the parallel component: \(v_\parallel = \sqrt{v^2 - v_\perp^2} = \sqrt{(5.0\times 10^7)^2 - (7.02\times 10^6)^2} = 4.87\times 10^7 m/s\)
04

Calculate the angle between the velocity and the magnetic field

Finally, we can determine the angle \(\theta\) between the velocity vector and the magnetic field using trigonometry. Using the definition of sine, we can write: \(\sin\theta = \frac{v_\perp}{v} = \frac{7.0\times 10^6}{5.0\times 10^7} = 0.1404\) Now, we find the inverse sine (arcsine) of this value to find the angle: \(\theta = \arcsin(0.1404) = 8.12^\circ\) To summarize: (a) The component of the positron's velocity perpendicular to the magnetic field is \(7.0\times 10^6 m/s\). (b) The component of the positron's velocity parallel to the magnetic field is \(4.87\times 10^7 m/s\). (c) The angle between the velocity and the magnetic field is \(8.12^\circ\).

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Most popular questions from this chapter

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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.
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