The magnitude of the magnetic force on the proton is 1.1 × 10⁻¹⁵ N
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Further explanation
Let's recall magnetic force on moving charge as follows:
[tex]\boxed{F = B q v \sin \theta}[/tex]
where:
F = magnetic force ( N )
B = magnetic field strength ( T )
q = charge of object ( C )
v = speed of object ( m/s )
θ = angle between velocity and direction of the magnetic field
Let's tackle the problem now !
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Given:
speed of proton = v = 4.0 × 10⁴ m/s
magnetic field strength = B = 0.20 T
charge of proton = q = 1.60 × 10⁻¹⁹ C
direction of speed = θ = 60°
Asked:
the magnitude of the magnetic force on the proton = F = ?
Solution:
[tex]F = B q v \sin \theta[/tex]
[tex]F = 0.20 \times 1.60 \times 10^{-19} \times 4.0 \times 10^4 \times \sin 60^o[/tex]
[tex]F = 6.4\sqrt{3} \times 10^{-16} \texttt{ N}[/tex]
[tex]\boxed{F \approx 1.1 \times 10^{-15} \texttt{ N}}[/tex]
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Learn more
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- Compare and contrast a series and parallel circuit : https://brainly.com/question/539204
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Answer details
Grade: High School
Subject: Physics
Chapter: Magnetic Field
