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.*
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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