A proton (charge = 1.6 x 10^-19 C and mass = 1.67 x 10^-27 kg) which is initially at rest, is accelerated through a potential difference of 300 V as shown in figure below. The speed of the proton as it exits the accelerator tube along the x-direction is [V] x 10^5 m/s. It then enters a region of uniform magnetic field of 0.1 T pointing in the positive z-direction (out of the page). The direction of the magnetic force on the proton as it enters this region of magnetic field is in the [D] direction. The magnitude of the magnetic force on the proton is [M] x 10^-15 N.
Select 3 correct answer(s)
[D] -y
[D] +y
[D] -z
[D] +z
[D] -x
[D] +x
[V] 2.4
[V] 4.5
[V] 3.6
[V] 1.2
[M] 3.84
[M] 5.23
[M] 2.67
[M] 1.32