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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

A uniform magnetic field of \(0.50 \mathrm{T}\) is directed to the north. At some instant, a particle with charge \(+0.020 \mu \mathrm{C}\) is moving with velocity \(2.0 \mathrm{m} / \mathrm{s}\) in a direction \(30^{\circ}\) north of east. (a) What is the magnitude of the magnetic force on the charged particle? (b) What is the direction of the magnetic force?
A solenoid has 4850 turns per meter and radius \(3.3 \mathrm{cm}\) The magnetic field inside has magnitude 0.24 T. What is the current in the solenoid?
The conversion between atomic mass units and kilograms is $$1 \mathrm{u}=1.66 \times 10^{-27} \mathrm{kg}$$ A sample containing carbon (atomic mass 12 u), oxygen \((16 \mathrm{u}),\) and an unknown element is placed in a mass spectrometer. The ions all have the same charge and are accelerated through the same potential difference before entering the magnetic field. The carbon and oxygen lines are separated by \(2.250 \mathrm{cm}\) on the photographic plate, and the unknown element makes a line between them that is \(1.160 \mathrm{cm}\) from the carbon line. (a) What is the mass of the unknown clement? (b) Identify the element.
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