Question

Suppose two parallel lines of current are 0.85A and 0.35A in the same direction, with the centers of their wires 21cm apart. A -2.04nC particle is fired at 3500m/s in the same direction as both lines of current exactly in between the lines of current.

a. Find the force that acts on the particle and in which direction. If a coordinate system will help, you may assume the currents and particle are in the +x direction and the currents lie in the xy plane.

b. Find the elctric field, magnitude, and direction that would allow the particle to continue to move parallel to the wires.

Thanks in advance! :) Please show all work!

Answer 1

The magnetic field of an infinitely long straight wire

where u0 = 4*pi*10^-7

Because of the two wires, magnetic field at the exactly in between the lines of current. = u0*I1 / ( 2*pi*r) + u0*I2 / ( 2*pi*r)

where r = 0.21/2 = 0.105 m

I1 = 0.85 A , I2 = 0.35 A

so, Magnetic field acting on charge = u0*I1 / ( 2*pi*r) + u0*I2 / ( 2*pi*r) = u0*(I1 + I2) / ( 2*pi*r)

Magnetic field acting on charge =   4*pi*10^-7*(0.85+ 0.35) / ( 2*pi*0.105) = 0.0000022857142 weber

Force on the charge = q*(V X B)

q = -2.04*10^-9 C

F = -2.04*10^-9 * ( 3500 * 0.0000022857142) = 16319.999 *10^-15 N

force that acts on the particle = 16319.999 *10^-15 N

Direction of force = the charge will experience a force towards one of the wires. perpendicular to it's inital direction.

the charge will experience a force in the y-direction.

b)

The force that requires t o be provided by electric field = 16319.999 *10^-15 N

so magnitude of Electric field E = Force / charge = (16319.999 *10^-15)/ ( 2.04*10^-9 ) = 0.0079999 N/C

magnitude of Electric field E = 0.0079999 N/C

direction of electric field would be in the y- direction opposite to the direction of force produced by the magnetic field .