Question

15125. Determine the force (magnitude and direction) on an a particle that is traveling at a speed of v = 9.50x10's in the plane of the paper and suddenly enters a uniform 5.04 T magnetic field that is oriented, as shown by the dots ( perpendicular to the paper. (Note: an a particle consists of 2 protons and two neutrons.) Answer OOOOOOOOO

Answer 1

Speed of alpha particle is not clear, if it is not $9.50\times 10^{7} m/s$ let me know in comments. I'll make proper corrections.

on a particle traveling in The force external Charged field magnetic 14 q (P x B are here the perpendiculara perpendicular magnetic field direction outwards &- Particle and The dot represents to paper. so j= y sox 10 m/s) R (5.04 ) 2 -19 2 X 166 x 10 Charge oh nettem x- particle q= so force Ť = <b>(16*18") [{C9.S0x10](33)} x {(5.40)(a)}] = (2) (1.6*10) (9.5x16)(5.40) (@x2) N f - (-9) N F = (1-6410 x 10'ON) (-9) Ve 1.6416 X10 -10 magnitude = 1.6416 X10 N direction downwards