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Chapter 19: Problem 29

The conversion between atomic mass units and kilograms is $$1 \mathrm{u}=1.66 \times 10^{-27} \mathrm{kg}$$ After being accelerated through a potential difference of \(5.0 \mathrm{kV},\) a singly charged carbon ion \(\left(^{12} \mathrm{C}^{+}\right)\) moves in a circle of radius \(21 \mathrm{cm}\) in the magnetic ficld of a mass spectrometer. What is the magnitude of the field?

Short Answer

Expert verified
Question: Calculate the magnitude of the magnetic field that causes a singly charged carbon ion to move in a circular path after being accelerated through a potential difference of 5.0 kV. The radius of the circular path is 21 cm. Answer: The magnitude of the magnetic field is 1.09 T.

Step by step solution

01

Identify given information

We are given: 1. Conversion between atomic mass units (u) and kilograms: 1 u = 1.66 * 10^(-27) kg 2. Potential difference (V) through which the ion is accelerated: 5.0 kV = 5.0 * 10^3 V 3. Mass of a singly charged carbon ion (m): 12 u = 12 * 1.66 * 10^(-27) kg 4. Radius of the circular path (r): 21 cm = 0.21 m 5. Charge of a singly charged carbon ion (q): +1 e = +1 * 1.60 * 10^(-19) C
02

Calculate the ion's final kinetic energy

Final kinetic energy: KE = qV We use the equation KE = qV, where KE is kinetic energy, q is the ion's charge, and V is the potential difference, to find the final kinetic energy of the ion. KE = (1 * 1.60 * 10^(-19) C)(5.0 * 10^3 V) = 8 * 10^(-16) J
03

Calculate the ion's final velocity

Relationship between kinetic energy and velocity: KE = (1/2)mv^2 We can rearrange the formula for kinetic energy, KE = (1/2)mv^2, to solve for the velocity (v): v = sqrt(2 * KE / m) where m is mass and KE is the kinetic energy we calculated in step 2. v = sqrt(2 * 8 * 10^(-16) J / (12 * 1.66 * 10^(-27) kg)) = 6.53 * 10^5 m/s
04

Calculate the magnetic field magnitude

Force acting on a charged particle in a magnetic field: F = qvB We can use the formula F = qvB, where F is the force acting on the ion, q is the ion's charge, v is the velocity, and B is the magnetic field magnitude. Since the ion is moving in a circular path due to the magnetic field, the centripetal force and magnetic force acting on it are equivalent: F_centr = F_magnetic => (mv^2) / r = qvB We can rearrange the formula to solve for the magnetic field magnitude (B): B = (mv^2) / (qr) where m is mass, v is the final velocity calculated in step 3, q is charge, and r is the radius of the circular path. B = ((12 * 1.66 * 10^(-27) kg)(6.53 * 10^5 m/s)^2) / ((1 * 1.60 * 10^(-19) C)(0.21 m)) = 1.09 T
05

Write the final answer

The magnitude of the magnetic field is 1.09 T.

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Most popular questions from this chapter

The strength of Earth's magnetic field, as measured on the surface, is approximately \(6.0 \times 10^{-5} \mathrm{T}\) at the poles and $3.0 \times 10^{-5} \mathrm{T}$ at the equator. Suppose an alien from outer space were at the North Pole with a single loop of wire of the same circumference as his space helmet. The diameter of his helmet is \(20.0 \mathrm{cm} .\) The space invader wishes to cancel Earth's magnetic field at his location. (a) What is the current required to produce a magnetic ficld (due to the current alone) at the center of his loop of the same size as that of Earth's field at the North Pole? (b) In what direction does the current circulate in the loop, CW or CCW, as viewed from above, if it is to cancel Earth's field?
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?
An electron moves at speed \(8.0 \times 10^{5} \mathrm{m} / \mathrm{s}\) in a plane perpendicular to a cyclotron's magnetic field. The magnitude of the magnetic force on the electron is \(1.0 \times 10^{-13} \mathrm{N}\) What is the magnitude of the magnetic field?
Electrons in a television's CRT are accelerated from rest by an electric field through a potential difference of \(2.5 \mathrm{kV} .\) In contrast to an oscilloscope, where the electron beam is deflected by an electric ficld, the beam is deflected by a magnetic field. (a) What is the specd of the electrons? (b) The beam is deflected by a perpendicular magnetic field of magnitude $0.80 \mathrm{T}$. What is the magnitude of the acceleration of the electrons while in the field? (c) What is the speed of the electrons after they travel 4.0 mm through the magnetic field? (d) What strength electric field would give the electrons the same magnitude acceleration as in (b)? (c) Why do we have to use an clectric ficld in the first place to get the electrons up to speed? Why not use the large acceleration due to a magnetic field for that purpose?
A magnet produces a \(0.30-\mathrm{T}\) field between its poles, directed to the east. A dust particle with charge \(q=-8.0 \times 10^{-18} \mathrm{C}\) is moving straight down at \(0.30 \mathrm{cm} / \mathrm{s}\) in this field. What is the magnitude and direction of the magnetic force on the dust particle?
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