Aproton travelling at 35 degree to an external magnetic field of magnitude 0.026 t experiences a force of magnitude 5.5x10^-17 n. hint: use formula f = qvbsineθ (a) calculate the speed of the proton. (b) calculate the kinetic energy of the proton in both joules (j) and electron volts (ev).

a) 2.3 x 10⁴ m/s

b) 4.42 x 10⁻¹⁹ J or 2.8 eV

Explanation:

θ = Angle between the direction of motion of proton and magnetic field = 35

B = magnitude of magnetic field = 0.026 T

F = magnitude of magnetic force =5.5 x 10⁻¹⁷ N

q = magnitude of charge on proton = 1.6 x 10⁻¹⁹ C

v = speed of proton

magnetic force on proton is given as

F = q v B Sinθ

5.5 x 10⁻¹⁷ = (1.6 x 10⁻¹⁹) (0.026) Sin35 v

v = 2.3 x 10⁴ m/s

b)

m = mass of the proton = 1.67 x 10⁻²⁷ kg

Kinetic energy of proton is given as

KE = (0.5) m v²

KE = (0.5) (1.67 x 10⁻²⁷) (2.3 x 10⁴)²

KE = 4.42 x 10⁻¹⁹ J

we also know that, 1 eV = 1.6 x 10⁻¹⁹ J

KE =

KE = 2.8 eV