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

speed

ratio of distance traveled to time interval

net force

vector sum of forces on object

object

source of diverging light rays either luminous or illuminated

don't remove my answer im trying my best to answer them

explanation: