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Question

Three coplanar parallel wires, each carrying a current of 10 A along the same direction, are placed with a separation 5.0 cm between the consecutive ones. Find the magnitude of the magnetic force per unit length acting on the wires.

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Solution

Let wires W1, W2 and W3 be arranged as shown in the figure.

Given:
Magnitude of current in each wire, i1 = i2 = i3 = 10 A
The magnetic force per unit length on a wire due to a parallel current-carrying wire is given by
Fl=μ0i1i22πd
So, for wire W1,
Fl =Flby wire W2+Flby wire W3 = μ0×10×102π×5×10-2+μ0×10×102π×10×10-2= 2×10-7×1025×10-2+2×10-7×10210×10-2= 4×10-4+2×10-4= 6×10-4 N

For wire W2,
Fl=Fl by wire W1-Fl by wire W3 = μ0×10×102π×5×10-2-μ0×10×102π×5×10-2 = 0

For wire W3,
Fl = Fl by wire W1+Fl by wire W2 =μ0×10×102π×5×10-2+μ0×10×102π×10×10-2
= 6 × 10−4 N

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