Item 3 Part A Midway between the wires. Two long, straight wires, one above the

Question

Item 3 Part A Midway between the wires. Two long, straight wires, one above the other, are seperated by a distance d 2.2 cm and are parallel to the x-axis. Let the ty-axis be in the plane of the wires in the direction from the lower wire to the upper wire. Each wire carries current 47 A in the +a - direction. Find the magnetic force on a negative point charge 25 C moving with velocity 2.58x 105 mis in the ty-direction, when the charge is: Enter your answers separated by commas. & Submit My Answers Give Up Part B at a distance d/2 cm above the upper wire. Enter your answers separated by commas. Submit My Answers Give Up Part C At a distance d/3 below the lower wire. Enter your answers separated by commas.

Explanation / Answer

Given

current in both the wires is i1=i2 = 47 A

separation of the wires is d = 2.2 cm = 0.022 m

point charge q = -25*10^-6 C

moving along +y direction with velocity v = 2.58*10^5 m/s

Part A

at midway between wires is

we know that the magnetic field at the center due to upper wire is out of the page and due to lower wire is into the page so the net field at center is zero so as the force  

F = B*q*v sin theta = 0 N

so the components of the forces are Fx = Fy= Fz =0 N

Part B

at distance d = d/2 cm above the upper wire is  

B1 due to upper wire is out of the page and due to lower wire is B2 also out of the page

B = B1+B2 = mue0*i1/(2pi*d/2) +mue0*i2/(2pi*(d+d/2))

B = (mue0*i/(2pid))(1/2 +3/2)

B = (4pi*10^-7*47/(2pi*0.022))(1/2+3/2) T = 0.0008545454545 T = 854.5454 *10^-6 T

now the forceis F = 854.5454 *10^-6 *25*10^-6 *2.58*10^5 sin90 N = 0.00551181783 N

and directed to the -ve x direction

so the components are  

Fx = -0.00551181783 N , Fy = 0 N , Fz = 0 N

Part C

at distance d/3 from lower wire

B1 due to upper wire is into the page and due to lower wire is B2 also into the page

B = B1+B2 = mue0*i1/(2pi*4d/3) +mue0*i2/(2pi*(d/3))

B = (mue0*i/(2pid))(4/3 +1/3)

B = (4pi*10^-7*47/(2pi*0.022))(4/3+1/3) T = 0.0007121212121 T = 712.1212 *10^-6 T

now the forceis F = 712.1212 *10^-6 *25*10^-6 *2.58*10^5 sin90 N = 0.004593182 N

directed to the +ve x direction so the Force components are  

Fx = 0.004593182 N , Fy = 0 N , Fz = 0 N

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