A particle with charge −
 5.20 nC
 is moving in a uniform magnetic field B⃗ =−(
 1.25 T
 )k^
. The magnetic force on the particle is measured to be F⃗ =−(
 4.00×10−7 N
 )i^+(
 7.60×10−7 N
 )j^
.
Part A
Part complete
Are there components of the velocity that are not determined by the measurement of the force?

yes
no
Previous Answers
 Correct
Part B
Calculate the x
-component of the velocity of the particle.
Express your answer in meters per second.
Activate to select the appropriates template from the following choices. Operate up and down arrow for selection and press enter to choose the input value typeActivate to select the appropriates symbol from the following choices. Operate up and down arrow for selection and press enter to choose the input value type

vx
 =
m/s
Previous AnswersRequest Answer
 Incorrect; Try Again; 5 attempts remaining
It seems that you have forgotten that the particle has a negative charge. Recall how the sign of the charge affects the direction of the magnetic force.
Part C
Calculate the y
-component of the velocity of the particle.
Express your answer in meters per second.
Activate to select the appropriates template from the following choices. Operate up and down arrow for selection and press enter to choose the input value typeActivate to select the appropriates symbol from the following choices. Operate up and down arrow for selection and press enter to choose the input value type

vy
 =
nothing
m/s


### Part A
Yes, there are components of the velocity that are not determined by the measurement of the force. Specifically, the component of the velocity in the direction of the magnetic field (the \( z \)-component in this case) cannot be determined from the magnetic force since the magnetic force is given by the cross product of velocity and the magnetic field. This force depends only on the perpendicular components of velocity relative to the magnetic field.

### Part B
To calculate the \( x \)-component of the velocity \( v_x \), we use the formula for the magnetic force:

\[
\vec{F} = q \vec{v} \times \vec{B}
\]

Given:
- \(\vec{F} = - (4.00 \times 10^{-7} \, \text{N}) \hat{i} + (7.60 \times 10^{-7} \, \text{N}) \hat{j}\)
- \( q = -5.20 \times 10^{-9} \, \text{C} \)
- \(\vec{B} = - (1.25 \, \text{T}) \hat{k} \)

The force components in the \( x \) and \( y \) directions are given by:

\[
F_x = q(v_y B_z - v_z B_y)
\]

\[
F_y = q(v_z B_x - v_x B_z)
\]

Since \( B \) is along the \( z \)-axis, the above equations simplify to:

\[
F_x = q v_y B_z
\]

\[
F_y = -q v_x B_z
\]

For the \( x \)-component of velocity:

\[
F_y = -q v_x B_z
\]

Solving for \( v_x \):

\[
v_x = -\frac{F_y}{q B_z}
\]

Substitute the values:

\[
v_x = -\frac{7.60 \times 10^{-7} \, \text{N}}{(-5.20 \times 10^{-9} \, \text{C}) \times (-1.25 \, \text{T})}
\]

Calculate \( v_x \):

\[
v_x = \frac{7.60 \times 10^{-7}}{6.5 \times 10^{-9}} \, \text{m/s} 
\]

\[
v_x \approx 1.169 \times 10^{2} \, \text{m/s}
\]

So, the \( x \)-component of the velocity is \( \boxed{117 \, \text{m/s}} \).

### Part C
For the \( y \)-component of velocity \( v_y \), use the equation:

\[
F_x = q v_y B_z
\]

Solving for \( v_y \):

\[
v_y = \frac{F_x}{q B_z}
\]

Substitute the values:

\[
v_y = \frac{-4.00 \times 10^{-7} \, \text{N}}{-5.20 \times 10^{-9} \, \text{C} \times (-1.25 \, \text{T})}
\]

Calculate \( v_y \):

\[
v_y = \frac{-4.00 \times 10^{-7}}{6.5 \times 10^{-9}} \, \text{m/s} 
\]

\[
v_y \approx -6.154 \times 10^{1} \, \text{m/s}
\]

So, the \( y \)-component of the velocity is \( \boxed{-61.5 \, \text{m/s}} \).

---

Would you like more details or have any other questions?

### Related Questions:
1. How is the force on a charged particle determined in an electric field?
2. What is the relationship between magnetic field direction and force on a moving charge?
3. How does changing the sign of the charge affect the direction of force in a magnetic field?
4. What are the physical implications if the particle had a positive charge instead?
5. How would the force change if the magnetic field was oriented along the \( x \)-axis instead?

**Tip:** Remember that the magnetic force is always perpendicular to both the velocity of the particle and the magnetic field.

Explore how to calculate the x-component and y-component of velocity for a charged particle moving in a magnetic field. This problem involves applying the magnetic force formula and understanding the interaction between velocity and magnetic field. Suitable for advanced high school students and beyond.

Math Problem Statement

vx

vy

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation