Question

Homework 5 Problem 20.90 11 of 11 Part A In (Figure 1) the top wire is 1.90-mm-diameter copper wire and is suspended in air due to the two magnetic forces from the bottom two wires. The current flow through the two bottom wires is 76 A in each. Calculate the required current flow in the suspended wire (M) Express your answer to two significant figures and include the appropriate units. IouValue Units Figure 1 of 1 Submit Request Answer Return to Assignment Provide Feedback CP 3.8 cm

Answer 1

(magnetic force on wire M due to wire N)

$F_P=\frac {\mu_0}{4\pi}\frac {2I_PI_M L}{r}$ (magnetic force on wire M due to wire P)

$F_P=\frac {\mu_0}{4\pi}\frac {2I_NI_M L}{r}$ ( $I_N=I_P$ )

$F_{net,mag}=\sqrt {F_N^2+F_M^2 +2F_NF_M\cos 60^{\degree}}$

$Mg=\sqrt {F_N^2+F_M^2 +2F_NF_M\cos 60^{\degree}}$ (Magnetic force is balanced by the gravity)

$\rho Vg=\sqrt {F_N^2+F_M^2 +2F_NF_M\cos 60^{\degree}}$

$\rho \pi R^2 Lg=\sqrt {F_N^2+F_M^2 +2F_NF_M\cos 60^{\degree}}$

$\rho \pi R^2 Lg=\sqrt 3\frac {\mu_0}{4\pi}\frac {2I_NI_M L}{r}$

$I_M=\frac {\rho \pi R^2 rg}{2\sqrt 3 \frac {\mu_0}{4\pi} I_N}$

$I_M=\frac {8920\cdot \pi (1.90\cdot 10^{-3})^2 \cdot 3.8\cdot 10^{-2}\cdot 9.8}{2\sqrt 3 \cdot 10^{-7} 76}$