Please refer again to the diagram. This time we have a charge of 5.52 x 10^-6 C moving at velocity 5.73 x 10^5 m/s through a magnetic field of strength 1.53 T and an electric field of strength 3.27 x 10^5 N/C. What is the magnitude of the total force on this charge? ### Diagram ExplanationThis diagram illustrates the motion of a charged particle in a region where both electric and magnetic fields are present.- **Electric Field (E):** Represented by vertical arrows pointing downward, indicates the direction and strength of the electric field. - **Magnetic Field (B):** Shown by 'X' symbols, indicating that the magnetic field is directed into the page.- **Particle Movement:** A charged particle (black dot) is moving with velocity \( v \), directed horizontally to the right.- **Forces Acting on the Particle:** - **Electric Force (\( F_e \)):** Points downward, parallel to the direction of the electric field. This is the force exerted by the electric field on the charged particle. - **Magnetic Force (\( F_m \)):** Points upward, perpendicular to both the velocity (\( v \)) and the magnetic field (B). This is the Lorentz force exerted by the magnetic field.### SummaryThe diagram represents a situation in which a charged particle is subjected to both electric and magnetic forces. The electric force pushes the particle downward, while the magnetic force pushes it upward, affecting the overall trajectory of the particle. This scenario is a common demonstration of Lorentz force in electromagnetic studies.

Given data: Charge (q) = 5.52×10-6 CVelocity (v) = 5.73×105 m/sMagnetic field (B) = 1.53 TElectric…

Answered: Please refer again to the diagram. This…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Please refer again to the diagram. This time we have a charge of 5.52 x 10^-6 C moving at velocity 5.73 x 10^5 m/s through a magnetic field of strength 1.53 T and an electric field of strength 3.27 x 10^5 N/C. What is the magnitude of the total force on this charge?

### Diagram Explanation

This diagram illustrates the motion of a charged particle in a region where both electric and magnetic fields are present.

- **Electric Field (E):** Represented by vertical arrows pointing downward, indicates the direction and strength of the electric field.

- **Magnetic Field (B):** Shown by 'X' symbols, indicating that the magnetic field is directed into the page.

- **Particle Movement:** A charged particle (black dot) is moving with velocity \( v \), directed horizontally to the right.

- **Forces Acting on the Particle:**
- **Electric Force (\( F_e \)):** Points downward, parallel to the direction of the electric field. This is the force exerted by the electric field on the charged particle.
- **Magnetic Force (\( F_m \)):** Points upward, perpendicular to both the velocity (\( v \)) and the magnetic field (B). This is the Lorentz force exerted by the magnetic field.

### Summary

The diagram represents a situation in which a charged particle is subjected to both electric and magnetic forces. The electric force pushes the particle downward, while the magnetic force pushes it upward, affecting the overall trajectory of the particle. This scenario is a common demonstration of Lorentz force in electromagnetic studies.

Expert Solution

Step 1: Determine the given variables

Given data:

Charge (q) = 5.52×10^-6 C

Velocity (v) = 5.73×10⁵ m/s

Magnetic field (B) = 1.53 T

Electric field (E) = 3.27×10⁵ N/C

Required:

The magnitude of the total force on the charge

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