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Chapter 27: Problem 29

An electron in a magnetic field moves counterclockwise on a circle in the \(x y\) -plane, with a cyclotron frequency of \(\omega=1.2 \cdot 10^{12} \mathrm{~Hz}\). What is the magnetic field, \(\vec{B}\) ?

Short Answer

Expert verified
Answer: The magnetic field acting on the electron is \(\vec{B} = 6.84 \times 10^{-4} \mathrm{T} \hat{z}\).

Step by step solution

01

Formula for cyclotron frequency

The formula for cyclotron frequency is given by: \(\omega = \frac{qB}{m}\) where \(\omega\) is the cyclotron frequency, \(q\) is the charge of the particle, \(B\) is the magnitude of the magnetic field, and \(m\) is the mass of the particle. For an electron, the charge \(q = -e\) and the mass \(m = m_e\), where \(e = 1.6 \times 10^{-19} \mathrm{C}\) and \(m_e = 9.11 \times 10^{-31} \mathrm{kg}\).
02

Rearranging the formula to find the magnitude of the magnetic field

Rearrange the formula to solve for the magnitude of the magnetic field \(B\): \(B = \frac{m\omega}{q}\)
03

Substitute the known values and calculate the magnetic field magnitude

Substitute the given value of \(\omega\) and known values of \(e\) and \(m_e\) into the formula and calculate \(B\): \(B = \frac{(9.11 \times 10^{-31} \mathrm{kg})(1.2 \times 10^{12} \mathrm{s}^{-1})}{1.6 \times 10^{-19} \mathrm{C}}\) After calculating, we get: \(B = 6.84 \times 10^{-4} \mathrm{T}\)
04

Determine the direction of the magnetic field

Here, as the electron moves in counterclockwise direction, the direction of the magnetic field \(\vec{B}\) must be perpendicular to the plane of its motion (xy-plane), along the positive z-direction. This follows the right-hand rule. So, the magnetic field \(\vec{B}\) is: \(\vec{B} = 6.84 \times 10^{-4} \mathrm{T} \hat{z}\)

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Most popular questions from this chapter

The velocity selector described in Solved Problem 27.2 is used in a variety of devices to produce a beam of charged particles of uniform velocity. Suppose the fields in such a selector are given by \(\vec{E}=\left(1.00 \cdot 10^{4} \mathrm{~V} / \mathrm{m}\right) \hat{x}\) and \(\vec{B}=(50.0 \mathrm{mT}) \hat{y} .\) Find the velocity in the \(z\) -direction with which a charged particle can travel through the selector without being deflected.

Show that the magnetic dipole moment of an electron orbiting in a hydrogen atom is proportional to its angular momentum, \(L: \mu=-e L / 2 m,\) where \(-e\) is the charge of the electron and \(m\) is its mass.

A coil consists of 120 circular loops of wire of radius \(4.8 \mathrm{~cm} .\) A current of 0.49 A runs through the coil, which is oriented vertically and is free to rotate about a vertical axis (parallel to the \(z\) -axis). It experiences a uniform horizontal magnetic field in the positive \(x\) -direction. When the coil is oriented parallel to the \(x\) -axis, a force of \(1.2 \mathrm{~N}\) applied to the edge of the coil in the positive \(y\) -direction can keep it from rotating. Calculate the strength of the magnetic field.

In which direction does a magnetic force act on an electron that is moving in the positive \(x\) -direction in a magnetic field pointing in the positive \(z\) -direction? a) the positive \(y\) -direction b) the negative \(y\) -direction c) the negative \(x\) -direction d) any direction in the \(x y\) -plane

A proton moving with a speed of \(4.0 \cdot 10^{5} \mathrm{~m} / \mathrm{s}\) in the positive \(y\) -direction enters a uniform magnetic field of \(0.40 \mathrm{~T}\) pointing in the positive \(x\) -direction. Calculate the magnitude of the force on the proton.
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