Question
A proton enters a magnetic field of intensity 1.5 Wb/m2 with a velocity 2×107 m/s in a direction at angle 30o with the field. The force on the proton will be (charge on proton is 1.6×10-19 C)
2.4×10-12 N
4.8×10-12 N
1.2×10-12 N
7.2×10-12 N
Solution
A.
2.4×10-12 N
Here :- q=1.6×10-19 C, B=1.5Wb/m2
v=2×107 m/s, θ=30o or Sin300 =
Force on proton is given by
F= q.v.BSinθ
=1.6 × 10-19 ×2 × 107 × 1.5 ×
F= 2.4×10-12 N
