A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magnitude of 3.73 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 2460 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 3.88 V/m
b. in the negative z direction and has a magnitude of 3.88 V/m
c. in the positive x direction and has a magnitude of 3.88 V/m

Explanation:

Force on moving proton due to magnetic field

F = BqV where B is magnetic field , q is charge on proton and v is its velocity

= 3.73 x 10⁻³ x 1.6 x 10⁻¹⁹ x 2460

= 14681.28 x 10⁻²²

= 14.68 x 10⁻¹⁹ N

According to Fleming's left hand rule , this force is towards positive z axis.

a ) electric field also acts in the positive z direction so , force on proton due to electric will be acting in positive z direction .

Magnitude of force = Eq where E is electric field and q is charge .

= 3.88 x 1.6 x 10⁻¹⁹ = 6.208 x 10⁻¹⁹ N

Net force will be sum of these two forces

= 14.68 x 10⁻¹⁹ N + 6.208 x 10⁻¹⁹

= 20.9 x 10⁻¹⁹ N

b )

electric field also acts in the negative  z direction so , force on proton due to electric will be acting in negative  z direction .

Magnitude of force = Eq where E is electric field and q is charge .

= 3.88 x 1.6 x 10⁻¹⁹ = 6.208 x 10⁻¹⁹ N

Net force will be difference  of these two forces

= 14.68 x 10⁻¹⁹ N -  6.208 x 10⁻¹⁹

= 8.5  x 10⁻¹⁹ N

c )

In the third case , no force will act due to magnetic field because magnetic field and velocity of proton are aligned on the same line .

Only force is due to electric field

Magnitude of force = Eq where E is electric field and q is charge .

= 3.88 x 1.6 x 10⁻¹⁹ = 6.208 x 10⁻¹⁹ N

This force will act along the electric field or along the x direction .

find the sum of the two horizontal vectors and the sum of the two vertical vectors

isoceles right angled triangle