### Problem Statement:A particle with mass \( 3.15 \times 10^{-4} \) kg carries a charge \( q \). The particle is given an initial horizontal velocity of \( 6.33 \times 10^4 \) m/s in the \( +x \) direction (the coordinate system is shown in the figure). The force of gravity acting on the particle tries to accelerate the particle downward, in the \( -y \) direction with gravitational acceleration \( g = 9.80 \) m/s\(^2\). A uniform magnetic field \( B = 1.46 \) T is applied in the \( +z \) direction (out of the page) such that the magnetic force balances out the gravitational force on the particle. What is the charge \( q \) of the particle (in Coulombs; indicate both the sign and magnitude)?### Figure Description:The figure illustrates the coordinate system:- The \( +y \) direction is pointing upwards.- The \( +x \) direction is pointing to the right.- The \( +z \) direction is represented as coming out of the page (denoted with a dot inside a circle).### Solution:To solve for the charge q of the particle, we use the balance of forces. The gravitational force \( F_g \) is given by:\[ F_g = mg \]The magnetic force \( F_m \) is given by:\[ F_m = qvB \]Since the magnetic force balances out the gravitational force, we have \( F_m = F_g \). Equating the two gives:\[ qvB = mg \]Rearranging to solve for \( q \):\[ q = \frac{mg}{vB} \]Substituting in the given values:- \( m = 3.15 \times 10^{-4} \) kg- \( g = 9.80 \) m/s\(^2\)- \( v = 6.33 \times 10^4 \) m/s- \( B = 1.46 \) T\[ q = \frac{(3.15 \times 10^{-4} \, \text{kg})(9.80 \, \text{m/s}^2)}{(6.33 \times 10^4 \, \text{m/s})(1.46 \, \

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(L09) A particle with mass 3.15E-4 kg carries a charge q. The particle is given an initial horizontal velocity 6.33E4 m/s in the +x direction (the coordinate system is shown in the figure). The force of gravity acting on the particle tries to accelerate the particle downward, in the -y direction with gravitational acceleration g = 9.80 m/s². A uniform magnetic field B = = 1.46 T is applied in +z direction (out of page) such that the magnetic force balance out the gravitational force on the particle. What is the charge q of the particle (in C; write down the sign and magnitude)? тул +ZO +x

(L09) A particle with mass 3.15E-4 kg carries a charge q. The particle is given an initial horizontal velocity 6.33E4 m/s in the +x direction (the coordinate system is shown in the figure). The force of gravity acting on the particle tries to accelerate the particle downward, in the -y direction with gravitational acceleration g = 9.80 m/s². A uniform magnetic field B = = 1.46 T is applied in +z direction (out of page) such that the magnetic force balance out the gravitational force on the particle. What is the charge q of the particle (in C; write down the sign and magnitude)? тул +ZO +x

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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Question

### Problem Statement:

A particle with mass \( 3.15 \times 10^{-4} \) kg carries a charge \( q \). The particle is given an initial horizontal velocity of \( 6.33 \times 10^4 \) m/s in the \( +x \) direction (the coordinate system is shown in the figure). The force of gravity acting on the particle tries to accelerate the particle downward, in the \( -y \) direction with gravitational acceleration \( g = 9.80 \) m/s\(^2\). A uniform magnetic field \( B = 1.46 \) T is applied in the \( +z \) direction (out of the page) such that the magnetic force balances out the gravitational force on the particle.

What is the charge \( q \) of the particle (in Coulombs; indicate both the sign and magnitude)?

### Figure Description:

The figure illustrates the coordinate system:
- The \( +y \) direction is pointing upwards.
- The \( +x \) direction is pointing to the right.
- The \( +z \) direction is represented as coming out of the page (denoted with a dot inside a circle).

### Solution:

To solve for the charge q of the particle, we use the balance of forces. The gravitational force \( F_g \) is given by:
\[ F_g = mg \]

The magnetic force \( F_m \) is given by:
\[ F_m = qvB \]

Since the magnetic force balances out the gravitational force, we have \( F_m = F_g \). Equating the two gives:
\[ qvB = mg \]

Rearranging to solve for \( q \):
\[ q = \frac{mg}{vB} \]

Substituting in the given values:
- \( m = 3.15 \times 10^{-4} \) kg
- \( g = 9.80 \) m/s\(^2\)
- \( v = 6.33 \times 10^4 \) m/s
- \( B = 1.46 \) T

\[ q = \frac{(3.15 \times 10^{-4} \, \text{kg})(9.80 \, \text{m/s}^2)}{(6.33 \times 10^4 \, \text{m/s})(1.46 \, \

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