Two wires parallel to each other and seperated by a distance d carry different c

Question

Two wires parallel to each other and seperated by a distance d carry different currents; they exert a force with magnitude F on each other.

(a) The wires carry current in the same direction. If the current carried by both wires is quadrupled and the distance between the wires is divided by five, then what will the force's magnitude be (as a multiple of F) and will it be repulsive or attractive?

(b) The wires carry current in opposite directions. If the current carried by one of the wires is quadrupled and the distance between the wires is divided by three, then what will the force's magnitude be (as a multiple of F) and will it be repulsive or attractive?

Explanation / Answer

Consider that First wire carries a current I1 and the second wire carries a current I2.
The second wire produces a magnetic field, B2 = oI2/2d.
The force on the first wire, f = B2 I1 x length
Force per unit length = f/length = B2I1
F = oI1 I2/2d.
Using the Fleming's left hand rule, we can see that if the direction of the current are same then the wires attract and repel if the directions are opposite.

a)
F = oI1 I2/2d.
I1 becomes 4I1, I2 becomes 4I2, d becomes d/5
Fnew = [o(4I1) x (4I2)]/2(d/5)
= 80 x [oI1 I2/2d]
= 80 F
Since the currents are in the same direction, they attract.

b)
I1 becomes 4I1
d becomes d/3
Fnew = [o(4I1) x (I2)]/2(d/3)
= 12 x [oI1 I2/2d]
= 12 F
Since the currents are in opposite direction, the wires repel.

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