Question

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)

Answer 1

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:

$F = B q v \sin \thetaF = 1.20 \times 1.60 \times 10^{-19} \times 4.0 \times 10^4 \times \sin 30^o\\{F =\ 3.8 \times 10^{-14} \texttt{ N}}$