Question

A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magnitude of 2.85 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 2840 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 5.38 V/m, (b) in the negative z direction and has a magnitude of 5.38 V/m, and (c) in the positive x direction and has a magnitude of 5.38 V/m

Answer 1

Step-by-step explanation:

a ) Magnetic force on proton

= B q v , B is magnetic field , q is charge with velocity v

= 2.85 x 10⁻³ x 1.6 x 10⁻¹⁹ x 2840

= 12.95 x 10⁻¹⁹ N

Its direction will be towards positive z - direction according to Fleming's left hand rule.

force on proton due to electric field = charge x electric field.

= 1.6 x 10⁻¹⁹ x 5.38

= 8.608 x 10⁻¹⁹ N

this force will be along the field ie in positive z direction so both the forces are acting in the same direction so they will add up.

total force = (12.95 + 8.608)x 10⁻¹⁹

= 21.558 x 10⁻¹⁹ N .

b ) in this case , both the forces are acting in opposite direction . net force

= (12.95 - 8.608)x 10⁻¹⁹

= 4.342 x 10⁻¹⁹ N

c ) In this case both the forces are acting perpendicular to each other

resultant = √(12.95² + 8.608²) x 10⁻¹⁹ N

= 15.54 x 10⁻¹⁹ N .