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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

Prove that the time for one revolution of a charged particle moving perpendicular to a uniform magnetic field is independent of its speed. (This is the principle on which the cyclotron operates.) In doing so, write an expression that gives the period \(T\) (the time for one revolution) in terms of the mass of the particle, the charge of the particle, and the magnetic field strength.
The torque on a loop of wire (a magnetic dipole) in a uniform magnetic field is \(\tau=N I A B\) sin \(\theta,\) where \(\theta\) is the angle between \(\overline{\mathbf{B}}\) and a line perpendicular to the loop of wire. Suppose an electric dipole, consisting of two charges \(\pm q\) a fixed distance \(d\) apart is in a uniform electric field \(\overrightarrow{\mathbf{E}}\). (a) Show that the net electric force on the dipole is zero. (b) Let \(\theta\) be the angle between \(\overrightarrow{\mathbf{E}}\) and a line running from the negative to the positive charge. Show that the torque on the electric dipole is \(\tau=q d E\) sin \(\theta\) for all angles $-180^{\circ} \leq \theta \leq 180^{\circ} .$ (Thus, for both electric and magnetic dipoles, the torque is the product of the dipole moment times the field strength times $\sin \theta .\( The quantity \)q d\( is the electric dipole moment; the quantity \)N I A$ is the magnetic dipole moment.)
An early cyclotron at Cornell University was used from the 1930 s to the 1950 s to accelerate protons, which would then bombard various nuclei. The cyclotron used a large electromagnet with an iron yoke to produce a uniform magnetic field of \(1.3 \mathrm{T}\) over a region in the shape of a flat cylinder. Two hollow copper dees of inside radius \(16 \mathrm{cm}\) were located in a vacuum chamber in this region. (a) What is the frequency of oscillation necessary for the alternating voltage difference between the dees? (b) What is the kinetic energy of a proton by the time it reaches the outside of the dees? (c) What would be the equivalent voltage necessary to accelerate protons to this energy from rest in one step (say between parallel plates)? (d) If the potential difference between the dees has a magnitude of $10.0 \mathrm{kV}$ each time the protons cross the gap, what is the minimum number of revolutions each proton has to make in the cyclotron?
An electromagnetic flowmeter is used to measure blood flow rates during surgery. Blood containing Na" ions flows due south through an artery with a diameter of \(0.40 \mathrm{cm} .\) The artery is in a downward magnetic field of \(0.25 \mathrm{T}\) and develops a Hall voltage of \(0.35 \mathrm{mV}\) across its diameter. (a) What is the blood specd (in \(\mathrm{m} / \mathrm{s}\) )? (b) What is the flow rate (in \(\mathrm{m}^{3} / \mathrm{s}\) )? (c) The leads of a voltmeter are attached to diametrically opposed points on the artery to measure the Hall voltage. Which of the two leads is at the higher potential?
You want to build a cyclotron to accelerate protons to a speed of $3.0 \times 10^{7} \mathrm{m} / \mathrm{s} .$ The largest magnetic field strength you can attain is 1.5 T. What must be the minimum radius of the dees in your cyclotron? Show how your answer comes from Newton's second law.
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