Find the magnitude and direction of the force (per unit length) that each wire e

Question

Find the magnitude and direction of the force (per unit length) that each wire experiences in Figure 21.57(b), using vector addition. I1 = 22.0 A, I2 = 11.00 A, and x = 27.0 cm.

*Angles are measured counterclockwise from the +x-axis.

Wire Magnitude Direction* A N/m ? B N/m ? C N/m ? D N/m ? Find the magnitude and direction of the force (per unit length) that each wire experiences in Figure 21.57(b), using vector addition. I1 = 22.0 A, I2 = 11.00 A, and x = 27.0 cm. Figure 21.57(b) *Angles are measured counterclockwise from the +x-axis.

Explanation / Answer

force per unit between parallel wires is given by F/L = uoi1i2/2pid

so here

F11 = 4*3.14e-7 * 22* 22/(2*3.14* 0.27)

F11 = 3.58 e-4 N/m along X axis

F12 = 4*3.14e-7 * 22* 22/(2*3.14* 0.27)

F12 = 3.58 e-4 N/m along y axis

as these two acts at right angles

Resulatant F' ^2 = 3.58^2 +3.58 ^2

F' = 5.06 e-4 N

F12 (diagonnaly) = uoi1i2/(2pid)

here d = sqrt(x^2+x^2)

d = sqrt(0.27^2 +0.27^2)

d = 0.381m

so

F12 Diag =    4*3.14e-7 * 22* 22/(2*3.14* 0.381)

F12 Diag = 2.54 e-4 N

so

angle between these two is tan theta = 2.54/5.06 = 0.501

theta = 26.65 deg

so

Fnet^2 = F1^2 +F2^2+ 2F1F2 cos theta

Fnet^2 = 5.06^2 + 2.54^2 + (2* 5.06 * 2.54 * cos 26.65)

Fnet^2 = 7.41*10^-4 N at 26.65 N on left top Cornor:

Similarly try other parts

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.