Answer:
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:
[tex]F = B q v \sin \thetaF = 1.20 \times 1.60 \times 10^{-19} \times 4.0 \times 10^4 \times \sin 30^o\\{F =\ 3.8 \times 10^{-14} \texttt{ N}}[/tex]
The magnitude of the force experienced by the proton (qp = 1.6 · 10-19 C) should be considered as the F =[tex]3.84 * 10^-14 N[/tex].
Since
Charge on proton = q = [tex]1.60 * 10^-19 C[/tex]
Velocity of proton = V =[tex]4.0 * 10^4 m/s[/tex]
Magnetic field = B = 0.20 T
Angle between V and B = θ = 60
Here,
F = magnetic force ( N )
B = magnetic field strength ( T )
q = charge of object ( C )
v = speed of object ( m/s )
So here the magnitude should be
[tex]= 1.20*1.60*10^-19 *4.0*10^4*sin 30\\\\= 3.8*10^-14 N[/tex]
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