A proton moving in the positive x direction with a speed of 9.7 x 105 m/s experiences zero magnetic force. When it moves in the positive y direction it experiences a force of 1.5 x 1013 N that points in the positive z direction. Determine the magnitude and direction of the magnetic field. magnitude

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### Problem Statement

A proton moving in the positive x direction with a speed of \(9.7 \times 10^5 \, \text{m/s}\) experiences zero magnetic force. When it moves in the positive y direction it experiences a force of \(1.5 \times 10^{-13} \, \text{N}\) that points in the positive z direction. Determine the magnitude and direction of the magnetic field.

#### Magnitude:
\[ \boxed{ \quad \text{T} \quad } \]

---

### Explanation

Given that:
- The proton's velocity in the x direction is \( 9.7 \times 10^5 \, \text{m/s} \) and it experiences zero magnetic force.
- The proton's velocity in the y direction is associated with a force of \( 1.5 \times 10^{-13} \, \text{N} \) in the positive z direction.

We need to determine the magnitude and direction of the magnetic field that causes these observations. 

When the proton moves in the x direction and experiences no magnetic force, it suggests that the magnetic field is also in the x direction. When the proton moves in the y direction and experiences a force in the z direction, it confirms the right-hand rule for the cross product of velocity and magnetic field vectors.

Using the fact that the magnetic force \( F \) on a charged particle is given by:
\[ F = q(v \times B) \]

Where:
- \( F \) is the force experienced by the proton
- \( q \) is the charge of a proton (\( 1.6 \times 10^{-19} \, \text{C} \))
- \( v \) is the velocity of the proton
- \( B \) is the magnetic field

Since the force is purely in the z direction and the proton's velocity is in the y direction, we can calculate the magnetic field magnitude \( B \).
Transcribed Image Text:### Problem Statement A proton moving in the positive x direction with a speed of \(9.7 \times 10^5 \, \text{m/s}\) experiences zero magnetic force. When it moves in the positive y direction it experiences a force of \(1.5 \times 10^{-13} \, \text{N}\) that points in the positive z direction. Determine the magnitude and direction of the magnetic field. #### Magnitude: \[ \boxed{ \quad \text{T} \quad } \] --- ### Explanation Given that: - The proton's velocity in the x direction is \( 9.7 \times 10^5 \, \text{m/s} \) and it experiences zero magnetic force. - The proton's velocity in the y direction is associated with a force of \( 1.5 \times 10^{-13} \, \text{N} \) in the positive z direction. We need to determine the magnitude and direction of the magnetic field that causes these observations. When the proton moves in the x direction and experiences no magnetic force, it suggests that the magnetic field is also in the x direction. When the proton moves in the y direction and experiences a force in the z direction, it confirms the right-hand rule for the cross product of velocity and magnetic field vectors. Using the fact that the magnetic force \( F \) on a charged particle is given by: \[ F = q(v \times B) \] Where: - \( F \) is the force experienced by the proton - \( q \) is the charge of a proton (\( 1.6 \times 10^{-19} \, \text{C} \)) - \( v \) is the velocity of the proton - \( B \) is the magnetic field Since the force is purely in the z direction and the proton's velocity is in the y direction, we can calculate the magnetic field magnitude \( B \).
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