shookiegriffin
10.03.2020 •
Physics
A proton is released such that it has an initial speed of 4.0 · 105 m/s from left to right across the page. A magnetic field of 1.2 T is present at an angle of 30° to the horizontal direction (or positive x axis). What is the magnitude of the force experienced by the proton? (qp = 1.6 · 10-19 C)
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Ответ:
Magnitude of the force on proton = F = 3.84 × 10^-14 N
Explanation:
Charge on proton = q = 1.60 × 10^-19 C
Velocity of proton = V = 4.0 × 10^4 m/s
Magnetic field = B = 0.20 T
Angle between V and B = θ = 60
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
Solution:
Ответ:
gdfurt yjfghetyje56 cfghdtthj gchnryh cgfnetdyj
Explanation: