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Chapter 28: Problem 2

In the Bohr model of the hydrogen atom, the electron moves in a circular orbit of radius \(5.3 \times 10^{-11} \mathrm{~m}\) with a speed of \(2.2 \times 10^{6} \mathrm{~m} / \mathrm{s}\). If we are viewing the atom in such a way that the electron's orbit is in the plane of the paper with the electron moving clockwise, find the magnitude and direction of the electric and magnetic fields that the electron produces at the location of the nucleus (treated as a point).

Short Answer

Expert verified
The magnitude of the electric field which the electron creates at the location of the nucleus would roughly be around \(5.12 脳 10^{11}\)N/C towards the electron. The magnitude of the magnetic field would be approximately \(6.21 脳 10^{-5}\)T directed out of the paper (since the electron is moving clockwise).

Step by step solution

01

Calculating the Electric Field

The electric field \(E\), in an atom due to charge \(q\) at a distance \(r\) is given by the formula \(E = k.e \frac{q}{r^2}\), where \(k.e\) is Coulomb's constant. Plugging in the values given in the problem (where \(r = 5.3 \times 10^{-11} m\), \(q = -e\), and \(k.e = 8.99 \times 10 ^{9} N.m ^2/C^2\), we get the magnitude of the electric field to be E = 8.99x\(10^9 \frac{N.m^2}{C^2}\) *\(\frac{1.6 \times 10^{-19} C}{(5.3 \times 10^{-11} m)^2}\)
02

Calculating the Magnetic Field

The magnetic field \(B\) due to a moving charge can be found using Ampere鈥檚 law, and is given by the equation \(B = \mu_0 \frac{I}{2 \pi r}\). Here, the current I can be expressed in terms of speed of electron and radius as, \(I = \frac{q}{T} = \frac{2\pi r}{T}\), where \(T= \frac{2\pi r}{v}\) is the time period of revolution. Inserting these values into the equation we derive the formula: \( B = \mu_0 \frac{v q }{2\pi r^2}\). Substitute \( \mu_0 = 4\pi\times 10^{-7} T.m/A, q = -e, v = 2.2 \times 10^6 m/s \) and \(r = 5.3 \times 10^{-11} m\).
03

Calculating the Direction of the Fields

The direction of the electric field at the nucleus caused by the negatively charged electron is towards the electron because electric fields point in the direction a positive test charge would be forced. The magnetic field is perpendicular to the plane of motion of the electron (right-hand thumb rule). Since the electron is moving clockwise, you would curl your right hand in the direction of the electron's movement. Your thumb then points out of the paper plane, indicating the direction of the magnetic field is out of the paper.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Field in an Atom
When we talk about the electric field within an atom, we are looking at the effects of one charged particle on another. In the case of a hydrogen atom, which has one electron orbiting around a single proton, the electric field plays a pivotal role. The electron, due to its charge, generates an electric field that influences the nucleus and vice versa.

Now, why is there an electric field? Every charged particle creates an electric field that extends out into the space around it. This field, invisible to the eye, exerts a force on other charged particles. For an electron in orbit, this force is vital because it's the electric force that keeps the electron bound to the nucleus and dictates its orbit.

The magnitude of this field diminishes with distance squared, precisely as indicated by the formula for the field due to a point charge: \(E = k.e \frac{q}{r^2}\). In our earlier example, the radius and charge determine how strong the electric field is at the nucleus.
Magnetic Field Due to a Moving Charge
Moving charges don't just create electric fields; they also create magnetic fields. This is why electric and magnetic fields are often linked, hence the term 'electromagnetism'.

An electron moving in a circular orbit, like in a hydrogen atom, is constantly changing direction. Due to this motion, it creates a magnetic field. This is essential for understanding how atoms interact with magnetic fields and why materials can be magnetic.

According to Ampere's law, the magnetic field created around a wire carrying current is directly proportional to the current and inversely proportional to the distance from the wire. While an atom is not a wire, the moving electron can be thought of as a tiny current. The magnetic field due to a moving charge in a wire or an electron in an atom can be calculated using modified versions of Ampere鈥檚 law nice and tidy encapsulated in the formula \(B = \mu_0 \frac{v q }{2\pi r^2}\).
Coulomb's Constant
Coulomb's constant (\(k.e\)), approximately \(8.99 \times 10^9 N.m^2/C^2\), is a cornerstone of electrostatics and appears in Coulomb's law.

This constant represents the electrostatic force per unit charge that two point charges exert on each other at a one-meter separation. It is crucial for determining the strength of the electric field created by a point charge. For example, when calculating the electric field at the nucleus of a hydrogen atom due to its single electron, as seen in the problem we've discussed, Coulomb's constant enables us to compute the exact magnitude of the electrostatic force involved.
Ampere's Law
Ampere's law relates the integrated magnetic field around a closed loop to the electric current passing through the loop. It's integral to our understanding of magnetism and is extensively used in electrical engineering and physics.

This law tells us how currents produce magnetic fields, which is not just theoretical 鈥� it's the principle that allows us to build motors and generators! In the context of our hydrogen atom problem, the moving electron generates a magnetic field due to its motion and this is described by a form of Ampere鈥檚 law. While this might sound a lot less mighty than the colossal structures of the industry, it鈥檚 the same principle whispering through both massive rotors and atomic orbits alike.

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Most popular questions from this chapter

Currents in dc transmission lines can be 100 A or higher. Some people are concerned that the electromagnetic fields from such lines near their homes could pose health dangers. For a line that has current \(150 \mathrm{~A}\) and a height of \(8.0 \mathrm{~m}\) above the ground, what magnetic field does the line produce at ground level? Express your answer in teslas and as a percentage of the earth's magnetic field, which is \(0.50 \mathrm{G}\). Is this value cause for worry?

The law of Biot and Savart in Eq. ( 28.7 ) generalizes to the case of surface currents as $$ \overrightarrow{\boldsymbol{B}}=\frac{\mu_{0}}{4 \pi} \int \frac{\sigma \overrightarrow{\boldsymbol{v}} \times \hat{\boldsymbol{r}}}{r^{2}} d a $$ where \(\sigma\) is the local charge density, \(\overrightarrow{\boldsymbol{v}}\) is the local velocity, and \(d a\) is a differential area element. Re-visit Challenge Problem 28.76 and use the above equation as an alternative means to derive the magnetic field at the center of the cylinder. Use the following steps: (a) Write the charge density \(\sigma\). (b) The origin is at the center of the cylinder. What is the vector \(\vec{v}\) that points from the element with coordinates \((x, y, z)=(x, R \cos \phi, R \sin \phi)\) to the origin? (c) What is the velocity \(\overrightarrow{\boldsymbol{v}}\) of the element? (d) What is the vector product \(\overrightarrow{\boldsymbol{v}} \times \hat{\boldsymbol{r}} ?\) (e) An area element on the cylinder may be written as \(d a=R d x d \phi .\) Use this and the previously established information to write the generalized law of Biot and Savart as a double integral. Evaluate the integral to determine the magnetic field \(\vec{B}\) at the center of the cylinder. (f) Is your result consistent with your result in Challenge Problem \(28.76 ?\)

Two concentric circular loops of wire lie on a tabletop, one inside the other. The inner wire has a diameter of \(20.0 \mathrm{~cm}\) and carries a clockwise current of \(12.0 \mathrm{~A}\), as viewed from above, and the outer wire has a diameter of \(30.0 \mathrm{~cm} .\) What must be the magnitude and direction (as viewed from above) of the current in the outer wire so that the net magnetic field due to this combination of wires is zero at the common center of the wires?

As a new electrical technician, you are designing a large solenoid to produce a uniform 0.150 T magnetic field near the center of the solenoid. You have enough wire for 4000 circular turns. This solenoid must be \(55.0 \mathrm{~cm}\) long and \(2.80 \mathrm{~cm}\) in diameter. What current will you need to produce the necessary field?

A closely wound, circular coil with radius \(2.40 \mathrm{~cm}\) has 800 turns. (a) What must the current in the coil be if the magnetic field at the center of the coil is \(0.0770 \mathrm{~T}\) ? (b) At what distance \(x\) from the center of the coil, on the axis of the coil, is the magnetic field half its value at the center?
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