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Chapter 19: Problem 95

(a) A proton moves with uniform circular motion in a magnetic field of magnitude 0.80 T. At what frequency \(f\) does it circulate? (b) Repeat for an electron.

Short Answer

Expert verified
Question: Calculate the frequency of rotation for a proton and an electron in a magnetic field of magnitude 0.80 T. Given the charge of a proton is \(1.60 \times 10^{-19} C\), mass of a proton is \(1.67 \times 10^{-27} kg\), charge of an electron is \(-1.60 \times 10^{-19} C\), and mass of an electron is \(9.11 \times 10^{-31} kg\). Answer: The frequency of rotation for the proton is approximately \(4.07 \times 10^7 Hz\) and for the electron, it is approximately \(-2.24 \times 10^{11} Hz\). The negative sign indicates that the electron rotates in the opposite direction as compared to the proton.

Step by step solution

01

Identify given parameters for a proton

Let's identify the given parameters for a proton. The charge of a proton (q) is \(1.60 \times 10^{-19} C\), the mass of a proton (m) is \(1.67 \times 10^{-27} kg\), and the magnetic field magnitude (B) is 0.80 T.
02

Use the formula to calculate the frequency for a proton

Now, we will use the formula \(f = \frac{q \cdot B}{2 \pi m}\) to calculate the frequency of rotation for a proton: \(f = \frac{(1.60 \times 10^{-19}\,C) \cdot (0.80\,T)}{2 \pi (1.67 \times 10^{-27}\,kg)}\) Calculate the value by solving the expression: \(f = \frac{1.28 \times 10^{-19}\,C\cdot T}{3.142 \times 10^{-27}\,kg} = \boxed{4.07 \times 10^7\,Hz}\) The frequency of rotation for the proton is approximately \(4.07 \times 10^7 Hz\). #b) Calculate the frequency of rotation for an electron#
03

Identify given parameters for an electron

Now let's identify the given parameters for an electron. The charge of an electron (q) is \(-1.60 \times 10^{-19} C\) (note the negative sign), the mass of an electron (m) is \(9.11 \times 10^{-31} kg\), and the magnetic field magnitude (B) remains the same at 0.80 T.
04

Use the formula to calculate the frequency for an electron

Now, we will use the formula \(f = \frac{q \cdot B}{2 \pi m}\) to calculate the frequency of rotation for an electron: \(f = \frac{(-1.60 \times 10^{-19}\,C) \cdot (0.80\,T)}{2 \pi (9.11 \times 10^{-31}\, kg)}\) Calculate the value by solving the expression: \(f = \frac{-1.28 \times 10^{-19}\,C\cdot T}{5.711 \times 10^{-31}\,kg} = \boxed{-2.24 \times 10^{11}\,Hz}\) The frequency of rotation for the electron is approximately \(-2.24 \times 10^{11} Hz\). The negative sign indicates that the electron rotates in the opposite direction as compared to the proton.

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Most popular questions from this chapter

The magnetic field in a cyclotron is \(0.50 \mathrm{T}\). What must be the minimum radius of the dees if the maximum proton speed desired is $1.0 \times 10^{7} \mathrm{m} / \mathrm{s} ?$
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