Two long straight wires carrying the same current I and separated by a distance r exert a force F on each other.If the current is increased to 4I and the separation is reduced to r/6 what is the...

Two long straight wires carrying the same current I and separated by a distance r exert a force F on each other.

If the current is increased to 4I and the separation is reduced to r/6 what is the force between two wires?

Asked on by fortunee

2 Answers | Add Yours

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

According to Ampere's law, the force between two wires carrying a current I1 and I2 and which are separated by a distance r is given as F= 2*K*I1*I2 / r, where K is a constant.

In the problem, when the wires carried a current I and were separated by a distance r the force was F

=> Fo = 2*K*I^2 / r

Now when the current is increased to 4I and the separation reduced to r/6, the new force is given by:

Fn = 2*K*(4I)^2 / (r/6)

=> 2*K*16*I^2*6/r

=> 2*K*I^2*(16*6)/r

=> Fo*(16*6)

=> Fo*(96)

Therefore the new force between the wires is 96 times the original force.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll write the formula of the force between 2 parallel wires:

F = (I1*I2/2pi*r)*no

In this probelm, the curent carried by the wires is the same, so, we'll modify the formula:

F = (I^2/2pi*r)*no

We'll re-write the formula if the current is increased by 4 times and separation is r/6.

F1 = (6*16I^2/2pi*r)*no

F1 = 96*(I^2/2pi*r)*no

But (I^2/2pi*r)*no = F and we'll substitute it:

F1 = 96*F

The increased force between 2 parallel wires is F1 = 96*F.

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