A proton enters a magnetic field of 2.5 Tesla with a speed of 1.5 x 10⁷ m/s at an angle of 30⁰ with the field. Find the force on the proton.
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Answer:
Angle between velocity of proton and magnetic field θ=30
o
Velocity of proton v=2×20
7
m/s
Magnetic field B=1.5Wb/m
2
Magnetic force acting on the proton F=qvBsinθ
∴ F=(1.6×10
−19
)(2×10
7
)(1.5)×0.5
⟹ F=2.4×10
−12
N
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