Question

Two high-current transmission lines carry currents of 29.0 A and 78.0 A in the same direction and are suspended parallel to each other 38.0 cm apart.Part A.) If the vertical posts supporting these wires divide the lines into straight 20.0m segments, what magnetic force does each segment exert on the other? (answer in F= ? N)Part B.) Is the force found in part (A) attractive or repulsive?

Answer 1

Answer with Explanation:

We are given that

$I_1=29 A$

$I_2=78 A$

$d=38 cm=\frac{38}{100}=0.38 m$

$1 m=100 cm$

a.Length of segment,l=20 m

Magnetic force ,F=$\frac{2\mu_0I_1I_2 l}{4\pi d}$

$\frac{\mu_0}{4\pi}=10^{-7}$

Substitute the values

$F=\frac{10^{-7}\times 29\times 78\times 20}{0.38}=0.0119 N$

Hence, the magnetic force exert by each segment on the other=0.0119 N

b.We know that when current carrying in the wires are in same direction then the force will attract to each other.

Hence, the force will be attractive.

Answer 2

Answer:

Explanation:

Current in one wire, I = 29 A

current in another wire, I' = 78 A

Distance between them, d = 38 cm = 0.38 m

length of wire, l = 20 m

(a) Force acting on the wires per unit length

$F=\frac{\mu _{0}}{4\pi }\frac{2II'}{d}$

$F=10^{-7}\times \frac{2\times 29\times 78}{0.38}$

F = 1.19 x 10^-3 N/m

total force on 20 m length

F' = F x 20 = 1.19 x 10^-3 x 20 = 0.0238 N

(b) As the current in both the wires in same direction so the force is attractive in nature .