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