Question

- Your answer is partially correct A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magnitude of 3.34 ml. At one Instant the velocity of the proton is in the positive y direction and has a magnitude of 2490 m/s. At that instant, what is the magnitude of the net force acting on the proton if the electric field is (a) in the positive z direction and has a magnitude of 3.96 V/m. (b) in the negative z direction and has a magnitude of 3.96 V/m, and (c) in the positive x direction and has a magnitude of 3.96 V/m? (a) Number i 1.33E-18 Units N (b) Number 1.96E-18 Units N (c) Number 2.37E-18 Units N

Answer 1

+y 6- 2490m)s - te E proton 3.34mT 3,964 Magnetic force acting on the proton F = 9ū x B) Fm = tex(ū x B) Fm = (1.6 x 10-'9) x (2490 X 3.34 x 10-?)X(+2) F = (13.307 x 10-19)x(+2) Electric force acting on the proton Fe=qĒ F, = teĚ Fe = (1.6 x 10-19) (3.96)x(+2) Fe = 6.336 x 10-19 x(+2) Magnitude of net force is F= Fm + F F=(13.307 x 10-'9) +(6.336 x 10-19) F= 19.643 x 10-19 N

ty 96 2490m)s X o te e proton 3.34mt 3,965 Magnetic force acting on the proton F = qū x B) F = tex(ū x B) F = (1.6 x 10-19) (2490 X 3.34 x 10-3)x(+2) Fm = (13.307 10-'9)x(+2) Electric force acting on the proton Fe=qĚ Fe = te Ē Fe = (1.6 x 10-19) (3.96)X(-2) Fe = 6.336 x 10-19 *(-2) Magnitude of net force is F= Fm - F F = (13.307 x 10-19) -(6.336 x 10-9) F=6.971 x 10-'' N

ty a u=2490m/s te È proton 3.34mT - 3,967 m Magnetic force acting on the proton Fm = f( ū x B) Fm = +ex(ū x B) Fm = (1.6 x 10-19) (2490 x 3.34 x 10-3)*(+2) F = (13.307 x 10-19) X (+2) Electric force acting on the proton Fe=qĒ F = te Ē F = (1.6 x 10-19) (3.96) X(+a) F = 6.336 x 10-19 X(+*) Magnitude of net force is F=VF. +F F= V(13.307 x 10-195? + (6.336 x 10-1972 F= 14.738 x 10-9 N