Question

The figure given below shows three straight long parallel conductors 1, 2 and 3 kept in x-y plane, carrying currents 2I, I and 3I respectively as shown in figure.

Find the magnitude and direction of:

1. net magnetic field at a point on conductor 1 and
2. net magnetic force acting on unit length of conductor 1, due to conductors 2 and 3.

Answer 1

i. Net Magnetic Field at Conductor 1:

Field due to conductor 2 (I along + `hat i`) at distance d:

Using the right-hand thumb rule,

`vec B_2 = (mu_0 I)/(2 pi d)hat k`   ...(Out of page)

Field due to conductor 3 (3I along `-hat i`) at distance 2d:

Using the right-hand thumb rule,

`vec B_3 = (mu_0(3 I))/(2 pi(2 d))(-hat k)`

= `(3 mu_0 I)/(4 pi d)(-hat k)`    ...(Into page)

Net field:

`vec B_"net" = vec B_2 + vec B_3`

= `((2 mu_0 I)/(4 pi d) - (3 mu_0 I)/(4 pi d)) hat k` 

= `-(mu_0 I)/(4 pi d) hat k`

Magnitude is `(mu_0 I)/(4 pi d)` and direction is into the page `(-hat k)`.

ii. Net Magnetic Force per Unit Length on Conductor 1:

Force due to 2 (F₁₂/L):

Currents are parallel (2I and I), so they attract.

`vec F_12//L = (mu_0(2 I)(I))/(2 pi d) (-hat j)`

Force due to 3 (F₁₃/L):

Currents are anti-parallel (2I and 3I), so they repel.

`vec F_13//L = (mu_0(2 I)(3 I))/(2 pi(2 d))(+hat j)`

= `(3 mu I^2)/(2 pi d)(+hat j)`

Net force `(vec F//L) = (mu_0 I^2)/(2 pi d) (-2 hat j + 3 hat j)`

= `(mu_0 I^2)/(2 pi d) hat j`