Question

Two long, straight wires, one above the other, are seperated by a distance d = 1.53 cm and are parallel to the x−axisx−axis. Let the +y−axis+y−axis be in the plane of the wires in the direction from the lower wire to the upper wire. Each wire carries current 34 A in the +x−direction+x−direction. Find the magnetic force on a negative point charge 79 μCμC moving with velocity 3.49×105 m/s in the +y-direction, when the charge is:

Answer 1

a)
`F_x = F_y = F_x = 0 \ \ N`

b)
`F_x= - 0.05718 \ N ; F_y = 0 \ N ; F_z= 0 \ N`

Step-by-step explanation:

Given that:

q = -79 μC

q = - 79×10⁻⁶ C

d = 1.53 cm

d = 0.0153 m

I = 34 A

v = 3.49×10⁵ m/s

`\bar{V} = v \hat{j}`

The force on charge q is given by

`\bar{ F } = q (\bar {v} * \bar{B})`

a) At midway (A) , the B will be :

`\bar{B} =\bar{B_1} + \bar{B_2}`

`B_1` and
`B_2` will be equally the same in respect to their magnitude but opposite direction at point A.

So;
`\bar = |B_2| = (\mu_oI)/(2 \pi d/2)`

where
`|B| =0`

∴
`\bar = 0`

∴
`F_x = F_y = F_x = 0 \ \ N`

b)

At a distance d/2 cm above the upper wire:

`\bar{B} =\bar{B_1} + \bar{B_2}`

where:

`B_1 =B_2 = (\mu_o I )/(2 \pi ( d + d/2 ))` ; upward to the plane of paper

∴
`(\mu_o I)/(2 \pi) \ \ [(2)/(3/d) +(2)/(d)] \ \ \hat{k}`

B =
`(\mu_o I)/( \pi \ d) \ \ [(1)/(3) +1] \ \ \hat{k}`

B =
`(\mu_o I)/( \pi \ d) \ [(4)/(3) ] \ \hat{k}`

B =
`(7 \pi * 10^(-7) *34)/(\pi*0.0153)*(4)/(3) \ \hat{k}`

B = 2.074 × 10⁻³T
`\hat {k}`

∴
`\bar {F} = q ( \bar{v} + \bar {B})`

`\bar {F} = q ( \bar{v} + B)j*k\\\\\bar {F} = q ( \bar{v} + B) \bar {i}`

`F = -79*10^(-6)*3.49*10^5*2.074*10^(-3)`

`F = - 0.05718 \ N \ \hat {i}`

`F_x= - 0.05718 \ N ; F_y = 0 \ N ; F_z= 0 \ N`