A charge q = +1.60E-12 C moves with speed 4.32E4 m/s through a constant magnetic field of strength 5.65 T. The charge's velocity vector makes an angle of 25.1° with the direction of the magnetic field. What is the magnetic force magnitude exerted on the charge (in Newtons (N))?
A charge q = +1.60E-12 C moves with speed 4.32E4 m/s through a constant magnetic field of strength 5.65 T. The charge's velocity vector makes an angle of 25.1° with the direction of the magnetic field. What is the magnetic force magnitude exerted on the charge (in Newtons (N))?
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![### Question 12:
A charge \( q = +1.60 \times 10^{-12} \, \text{C} \) moves with a speed of \( 4.32 \times 10^{4} \, \text{m/s} \) through a constant magnetic field of strength \( 5.65 \, \text{T} \). The charge's velocity vector makes an angle of \( 25.1^\circ \) with the direction of the magnetic field. What is the magnetic force magnitude exerted on the charge (in Newtons \(\text{N}\))?
### Explanation:
To solve for the magnetic force \( F \) exerted on the charge, we can use the formula:
\[ F = qvB \sin(\theta) \]
where:
- \( q \) is the charge,
- \( v \) is the speed of the charge,
- \( B \) is the magnetic field strength,
- \( \theta \) is the angle between the velocity vector and the magnetic field.
Given:
- \( q = 1.60 \times 10^{-12} \, \text{C} \)
- \( v = 4.32 \times 10^{4} \, \text{m/s} \)
- \( B = 5.65 \, \text{T} \)
- \( \theta = 25.1^\circ \)
Substituting these values into the formula:
\[ F = (1.60 \times 10^{-12} \, \text{C})(4.32 \times 10^{4} \, \text{m/s})(5.65 \, \text{T}) \sin(25.1^\circ) \]
Calculate \( \sin(25.1^\circ) \):
\[ \sin(25.1^\circ) \approx 0.424 \]
Then the force \( F \) is:
\[ F = (1.60 \times 10^{-12})(4.32 \times 10^{4})(5.65)(0.424) \]
\[ F \approx (1.60 \times 4.32 \times 5.65 \times 0.424) \times 10^{-12} \]
\[ F \approx (1.60 \times 4.32 \times 5.65 \times](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb234898a-ca00-4740-ae3d-79576f672843%2F6011da51-ac3f-4f1b-9436-8fd74129fafb%2Far6n3g_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Question 12:
A charge \( q = +1.60 \times 10^{-12} \, \text{C} \) moves with a speed of \( 4.32 \times 10^{4} \, \text{m/s} \) through a constant magnetic field of strength \( 5.65 \, \text{T} \). The charge's velocity vector makes an angle of \( 25.1^\circ \) with the direction of the magnetic field. What is the magnetic force magnitude exerted on the charge (in Newtons \(\text{N}\))?
### Explanation:
To solve for the magnetic force \( F \) exerted on the charge, we can use the formula:
\[ F = qvB \sin(\theta) \]
where:
- \( q \) is the charge,
- \( v \) is the speed of the charge,
- \( B \) is the magnetic field strength,
- \( \theta \) is the angle between the velocity vector and the magnetic field.
Given:
- \( q = 1.60 \times 10^{-12} \, \text{C} \)
- \( v = 4.32 \times 10^{4} \, \text{m/s} \)
- \( B = 5.65 \, \text{T} \)
- \( \theta = 25.1^\circ \)
Substituting these values into the formula:
\[ F = (1.60 \times 10^{-12} \, \text{C})(4.32 \times 10^{4} \, \text{m/s})(5.65 \, \text{T}) \sin(25.1^\circ) \]
Calculate \( \sin(25.1^\circ) \):
\[ \sin(25.1^\circ) \approx 0.424 \]
Then the force \( F \) is:
\[ F = (1.60 \times 10^{-12})(4.32 \times 10^{4})(5.65)(0.424) \]
\[ F \approx (1.60 \times 4.32 \times 5.65 \times 0.424) \times 10^{-12} \]
\[ F \approx (1.60 \times 4.32 \times 5.65 \times
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