Two protons (charge q= 1.602·10-19 C) move at the same speed v= 1.4 ·106 m/s in opposite directions parallel to the x-axis. At the instant when they are at the same x-position, the proton moving in the negative direction is at distance r= 0.7 mm in the positive y-direction with respect to the one moving in the positive direction.
What is the magnitude of the magnetic field at the point of the proton moving in the negative direction?
B=
What is the magnetic force direction experienced by the proton moving in the negative direction?
FB=

(a) The magnitude of the magnetic field at the point of the proton moving in the negative direction is 2.1 x 10⁻⁹ T.(b) The magnetic force direction experienced by the proton moving in the negative direction is 4.71 x 10⁻²² N.What is the magnetic field due to the proton?The magnitude of the magnetic field at the point of the proton moving in the negative direction is calculated as follows;F = qvB = kq²/r²qvB = kq²/r²B = kq²/qvr²B = kq/vr²where;k is coulomb's constantq is the magnitude of charge of the protonsr is the distance between the protonsv is the speed of the protonB = (9 x 10⁹ x 1.602 x 10⁻¹⁹) / (1.4 x 10⁶ x 0.7 x 10⁻³ x 0.7 x 10⁻³ )B = 2.1 x 10⁻⁹ TThe magnetic force direction experienced by the proton moving in the negative direction is calculated as follows;F = qvBF = 1.602 x 10⁻¹⁹ x 1.4 x 10⁶ x  2.1 x 10⁻⁹F = 4.71 x 10⁻²² N.Learn more about magnetic force here: https://brainly.com/question/13277365#SPJ1