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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

It would be mathematically possible, for a region with zero current density, to define a scalar magnetic potential analogous to the electrostatic potential: \(V_{B}(\vec{r})=-\int_{\vec{r}_{0}}^{\vec{r}} \vec{B} \cdot d \vec{s},\) or \(\vec{B}(\vec{r})=-\nabla V_{B}(\vec{r}) .\) However, this has not been done. Explain why not.

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.

A copper wire with density \(\rho=8960 \mathrm{~kg} / \mathrm{m}^{3}\) is formed into a circular loop of radius \(50.0 \mathrm{~cm} .\) The cross-sectional area of the wire is \(1.00 \cdot 10^{-5} \mathrm{~m}^{2},\) and a potential difference of \(0.012 \mathrm{~V}\) is applied to the wire. What is the maximum angular acceleration of the loop when it is placed in a magnetic field of magnitude \(0.25 \mathrm{~T}\) ? The loop rotates about an axis through a diameter.

A semicircular loop of wire of radius \(R\) is in the \(x y\) -plane, centered about the origin. The wire carries a current, \(i\), counterclockwise around the semicircle, from \(x=-R\) to \(x=+R\) on the \(x\) -axis. A magnetic field, \(\vec{B}\), is pointing out of the plane, in the positive \(z\) -direction. Calculate the net force on the semicircular loop.

A circular coil with a radius of \(10.0 \mathrm{~cm}\) has 100 turns of wire and carries a current, \(i=100 . \mathrm{mA} .\) It is free to rotate in a region with a constant horizontal magnetic field given by \(\vec{B}=(0.0100 \mathrm{~T}) \hat{x}\). If the unit normal vector to the plane of the coil makes an angle of \(30.0^{\circ}\) with the horizontal, what is the magnitude of the net magnetic torque acting on the coil? 27.61 At \(t=0\) an electron crosses the positive \(y\) -axis (so \(x=0\) ) at \(60.0 \mathrm{~cm}\) from the origin with velocity \(2.00 \cdot 10^{5} \mathrm{~m} / \mathrm{s}\) in the positive \(x\) -direction. It is in a uniform magnetic field. a) Find the magnitude and the direction of the magnetic field that will cause the electron to cross the \(x\) -axis at \(x=60.0 \mathrm{~cm}\). b) What work is done on the electron during this motion? c) How long will the trip take from \(y\) -axis to \(x\) -axis?
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