A proton, with a charge Q = 1.602 x 10-19 Coulombs has an initial velocity as shown below with a magnitude of v = The proton moves in the presence of a 4.0 Tesla uniform magnetic field going into the page, with a velocity always perpendicular to the magnetic field. %3D 2.0 x 10' m/s. Take the proton mass to be M = 1.67 × 10-27 kg.
A proton, with a charge Q = 1.602 x 10-19 Coulombs has an initial velocity as shown below with a magnitude of v = The proton moves in the presence of a 4.0 Tesla uniform magnetic field going into the page, with a velocity always perpendicular to the magnetic field. %3D 2.0 x 10' m/s. Take the proton mass to be M = 1.67 × 10-27 kg.
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![**Title: Proton Motion in Magnetic Field**
A proton, with a charge \( Q = 1.602 \times 10^{-19} \) Coulombs, has an initial velocity as shown below with a magnitude of \( v = 2.0 \times 10^7 \) m/s. Take the proton mass to be \( M = 1.67 \times 10^{-27} \) kg. The proton moves in the presence of a 4.0 Tesla uniform magnetic field, going into the page, with a velocity always perpendicular to the magnetic field.
**Diagram Explanation:**
- A series of "X"s represent the magnetic field going into the page.
- A circular path indicates the trajectory of the proton.
- A "+" denotes the position of the proton, moving perpendicular to the magnetic field shown by an arrow (rightward).
**Proton Velocity in Relation to Magnetic Field**
a. **Sketch:** On the figure, the subsequent (circular) path followed by the proton is shown.
b. **Calculate the Radius:**
Formula:
\[
R = \frac{mv}{QB}
\]
Substitution:
\[
R = \frac{(1.67 \times 10^{-27})(2 \times 10^7)}{(1.602 \times 10^{-19})(4)}
\]
Result:
\[
R = 5.1 \times 10^{-8} \, \text{m}
\]
c. **Calculate the Frequency of Orbits:**
Formula:
\[
f = \frac{QB}{2\pi M}
\]
Substitution:
\[
f = \frac{(1.602 \times 10^{-19})(4)}{2\pi(1.67 \times 10^{-27})}
\]
Result:
\[
f = 1.60 \times 10^{-55} \, \text{sec}^{-1}
\]
*Note: Mistaken rounding in scientific notation. Modify as necessary.*](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa442f039-c23f-4767-9723-571e7d7e1a22%2F30cdfc60-e39c-4a9f-98df-2f62001746c8%2Fvz7fri.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Proton Motion in Magnetic Field**
A proton, with a charge \( Q = 1.602 \times 10^{-19} \) Coulombs, has an initial velocity as shown below with a magnitude of \( v = 2.0 \times 10^7 \) m/s. Take the proton mass to be \( M = 1.67 \times 10^{-27} \) kg. The proton moves in the presence of a 4.0 Tesla uniform magnetic field, going into the page, with a velocity always perpendicular to the magnetic field.
**Diagram Explanation:**
- A series of "X"s represent the magnetic field going into the page.
- A circular path indicates the trajectory of the proton.
- A "+" denotes the position of the proton, moving perpendicular to the magnetic field shown by an arrow (rightward).
**Proton Velocity in Relation to Magnetic Field**
a. **Sketch:** On the figure, the subsequent (circular) path followed by the proton is shown.
b. **Calculate the Radius:**
Formula:
\[
R = \frac{mv}{QB}
\]
Substitution:
\[
R = \frac{(1.67 \times 10^{-27})(2 \times 10^7)}{(1.602 \times 10^{-19})(4)}
\]
Result:
\[
R = 5.1 \times 10^{-8} \, \text{m}
\]
c. **Calculate the Frequency of Orbits:**
Formula:
\[
f = \frac{QB}{2\pi M}
\]
Substitution:
\[
f = \frac{(1.602 \times 10^{-19})(4)}{2\pi(1.67 \times 10^{-27})}
\]
Result:
\[
f = 1.60 \times 10^{-55} \, \text{sec}^{-1}
\]
*Note: Mistaken rounding in scientific notation. Modify as necessary.*
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