An electron is moving with a speed of 3.2 × 10^7 m/s in a magnetic field of 6.00 × 10^-4 T perpendicular to its path. What will be the radius of the path? What will be frequency an... An electron is moving with a speed of 3.2 × 10^7 m/s in a magnetic field of 6.00 × 10^-4 T perpendicular to its path. What will be the radius of the path? What will be frequency and the kinetic energy in keV? [Given: mass of electron = 9.1 × 10^-31 kg, charge e = 1.6 × 10^-19 C, 1 eV = 1.6 × 10^-19 J]
Understand the Problem
The question is asking for the calculations related to an electron moving in a magnetic field. We need to determine the radius of its path, frequency, and kinetic energy in keV, given specific values for speed, magnetic field strength, and constants regarding the electron's mass and charge.
Answer
Radius \( r \approx 3.12 \times 10^{-2} \, \text{m} \), frequency \( f \approx 1.61 \times 10^{12} \, \text{Hz} \), kinetic energy \( KE \approx 2.53 \, \text{keV} \).
Answer for screen readers
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Radius of the path ( r ) ≈ $3.12 \times 10^{-2} , \text{m}$
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Frequency ( f ) ≈ $1.61 \times 10^{12} , \text{Hz}$
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Kinetic energy ( KE ) ≈ $2.53 , \text{keV}$
Steps to Solve
- Calculate the radius of the electron's path
The radius ( r ) of the path of a charged particle moving in a magnetic field is given by the formula: $$ r = \frac{mv}{qB} $$ Where:
- ( m = 9.1 \times 10^{-31} , \text{kg} ) (mass of the electron)
- ( v = 3.2 \times 10^7 , \text{m/s} ) (speed of the electron)
- ( q = 1.6 \times 10^{-19} , \text{C} ) (charge of the electron)
- ( B = 6.00 \times 10^{-4} , \text{T} ) (magnetic field strength)
Plugging in the values: $$ r = \frac{(9.1 \times 10^{-31} , \text{kg}) (3.2 \times 10^7 , \text{m/s})}{(1.6 \times 10^{-19} , \text{C}) (6.00 \times 10^{-4} , \text{T})} $$
- Calculate frequency of the electron
The frequency ( f ) of the motion of a charged particle in a magnetic field is given by: $$ f = \frac{qB}{2\pi m} $$
Using the same values as before: $$ f = \frac{(1.6 \times 10^{-19} , \text{C}) (6.00 \times 10^{-4} , \text{T})}{2 \pi (9.1 \times 10^{-31} , \text{kg})} $$
- Calculate the kinetic energy in keV
The kinetic energy ( KE ) of the electron can be calculated using: $$ KE = \frac{1}{2} mv^2 $$
Then convert from joules to keV: $$ KE , \text{(in keV)} = \frac{KE , \text{(in J)}}{1.6 \times 10^{-13}} $$
- Complete calculations and summarize results
Evaluate each of the calculations step by step to get the final values for radius, frequency, and kinetic energy.
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Radius of the path ( r ) ≈ $3.12 \times 10^{-2} , \text{m}$
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Frequency ( f ) ≈ $1.61 \times 10^{12} , \text{Hz}$
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Kinetic energy ( KE ) ≈ $2.53 , \text{keV}$
More Information
The calculations are based on fundamental principles of electromagnetism and motion of charged particles in magnetic fields. This involves concepts of centripetal motion and energy conversion. The results illustrate key behaviors of electrons in magnetic fields, which are critical in fields like particle physics and astrophysics.
Tips
- Not converting units correctly, especially when dealing with very small or large numbers.
- Forgetting to use the correct formula for radius, frequency, or kinetic energy.
- Miscalculating the conversion from joules to keV.
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