Question

a proton of energy 3.4MeV moves vertically downwards through a horizontal magnetic field of 3T which acts from South to North what is the force of the proton mass of proton is 1.7 x 10^- 27 kg charge on proton is 1.6x 10^- 19

Answer 1

Answer:

F = qvBsinθ

given, q = 1.6 x 10⁻¹⁹C , B = 3T, v = ?

sinθ = 1 (∵velocity and B are perpendicular to each other)  

1eV = 1.6 x 10⁻¹⁹ J

KE = 3.4MeV

using KE to find v, we use v = $\sqrt{\frac{2KE}{m}}$

F = ($qB\sqrt{\frac{2KE}{m}}$)

F = $1.6*10^{-19}C*3T*\sqrt{\frac{2*3.4*10^6*1.6*10^{-19}J}{1.7*10^{-27}kg}}$

F = 1.214 x 10⁻¹¹ N