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Chapter 19: Problem 88

In a simple model, the electron in a hydrogen atom orbits the proton at a radius of \(53 \mathrm{pm}\) and at a constant speed of $2.2 \times 10^{6} \mathrm{m} / \mathrm{s} .$ The orbital motion of the electron gives it an orbital magnetic dipole moment. (a) What is the current \(I\) in this current loop? [Hint: How long does it take the electron to make one revolution?] (b) What is the orbital dipole moment IA? (c) Compare the orbital dipole moment with the intrinsic magnetic dipole moment of the electron $\left(9.3 \times 10^{-24} \mathrm{A} \cdot \mathrm{m}^{2}\right)$

Short Answer

Expert verified
Solution: (a) First, find the current I in the current loop by calculating the time taken for the electron to make one revolution and dividing the elementary charge by time. (b) Next, find the orbital dipole moment IA by multiplying the current obtained in part (a) with the area of the current loop (c) Finally, compare the values of the orbital dipole moment, IA, with the given intrinsic magnetic dipole moment of the electron, which is \(9.3 \times 10^{-24} \mathrm{A} \cdot \mathrm{m}^{2}\).

Step by step solution

01

(a) Finding the current I in the current loop

First, we need to find the time taken for the electron to make one full orbit. To do that, we use the formula for the circumference of a circle and divide it by the electron's speed: Circumference = \(2\pi r\) where r is the radius of the orbit. Time taken for one revolution, T = Circumference / Speed T = \(\frac{2\pi(53 \times 10^{-12} m)}{2.2 \times 10^6 m/s}\) Now, we'll find the current using the formula: Current, I = \(\frac{Charge}{Time}\) where Charge is the elementary charge of the electron (e = \(1.6 \times 10^{-19} C\)). I = \(\frac{1.6 \times 10^{-19}C}{T}\)
02

(b) Finding the orbital dipole moment IA

The orbital dipole moment, IA, can be calculated using the current obtained in part (a) and the area of the current loop (the circle formed by the electron's movement). The formula for the area, A, of a circle is: A = \(\pi r^{2}\) So, the orbital dipole moment is: IA = \(I \times \pi (53 \times 10^{-12})^{2} m^{2}\)
03

(c) Comparing the orbital dipole moment with the intrinsic magnetic dipole moment of the electron

Now, we simply compare the value of the orbital dipole moment, IA, obtained in part (b) with the given intrinsic magnetic dipole moment of the electron, which is \(9.3 \times 10^{-24} \mathrm{A} \cdot \mathrm{m}^{2}\). This will let us see how the orbital magnetic dipole moment is related to the intrinsic magnetic dipole moment of the electron.

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Most popular questions from this chapter

Two identical bar magnets lic onatable along a straight line with their north poles facing each other. Sketch the magnetic field lines.
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?
The magnetic field in a cyclotron is \(0.50 \mathrm{T}\). Find the magnitude of the magnetic force on a proton with speed $1.0 \times 10^{7} \mathrm{m} / \mathrm{s}$ moving in a plane perpendicular to the field.
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.
An electron moves with speed \(2.0 \times 10^{5} \mathrm{m} / \mathrm{s}\) in a uniform magnetic field of \(1.4 \mathrm{T},\) pointing south. At one instant, the electron experiences an upward magnetic force of $1.6 \times 10^{-14} \mathrm{N} .$ In what direction is the electron moving at that instant? 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 possibilitics.)
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