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Paladinen [302]
3 years ago
5

In a microwave oven, electrons describe circular motion in a magnetic field within a special tube called amagnetron; as you'll l

earn in Chapter 29, the electrons' motion results in the production of micowaves. The electrons circle at a frequency of 2.15 GHz . The magnetron can accommodate electron orbits with maximum diameter 2.62 mm .
Part A

What's the magnetic field strength?

Part B

What's the electrons' energy in eV.
Physics
1 answer:
QveST [7]3 years ago
5 0

Answer:

The magnetic field strength and the electrons' energy are 0.077 T and 0.8906 eV.

Explanation:

Given that,

Diameter = 2.62 mm

Frequency = 2.15 GHz

(A). We need to calculate the magnetic field strength

Using formula of the magnetic field strength

B=\dfrac{2\pi mf}{e}

Where, f = frequency

e = charge of electron

Put the value into the formula

B=\dfrac{2\times3.14\times9.1\times10^{-31}\times2.15\times10^{9}}{1.6\times10^{-19}}

B=0.077\ T

(B). We need to calculate the energy of electron

Using formula of energy

E=\dfrac{1}{2}m(r\omega)^2

E=\dfrac{1}{2}\times9.1\times10^{-31}\times(1.31\times10^{-3}\times2\pi\times2.15\times10^{9})^2

E=1.4249\times10^{-16}\ J

The energy in eV

1 eV=1.6\times10^{-16}\ J

E=\dfrac{1.4249\times10^{-16}}{1.6\times10^{-16}}

E=0.8906\ eV

Hence, The magnetic field strength and the electrons' energy are 0.077 T and 0.8906 eV.

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The representation of this problem is shown in Figure 1. So our goal is to find the vector \overrightarrow{R}. From the figure we know that:

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From geometry, we know that:

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So if you want to find out <span>how far are you from your starting point you need to know the magnitude of the vector \overrightarrow{R}, that is:
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\left | \overrightarrow{R} \right |=&#10;\sqrt{R_x^2+R_y^2}=\sqrt{(-19.42)^2+(-1.58)^2}=\boxed{19.48m}

Finally, let's find the <span>compass direction of a line connecting your starting point to your final position. What we are looking for here is an angle that is shown in Figure 2 which is an angle defined with respect to the positive x-axis. Therefore:

</span>\theta_R=180^{\circ}+tan^{-1}(\frac{\left | R_y \right |}{\left | R_x \right |}) \\ \\ \theta_R=180^{\circ}+tan^{-1}(\frac{1.58}{19.42}) \\ \\ \theta_R=180^{\circ}+4.65^{\circ}=185.85^{\circ}


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