(a) Find the resultant magnetic force exertedby the long wire on the square curr

Question

(a) Find the resultant magnetic force exertedby the long wire on the square current loop in the figure below ifI1 = 7.76A and I2 = 4.16 A. The edge length of the square is18.2 cm, and the distance from the longwire to the closest edge of the loop is 4.50 cm. *I have gotten all the answers but the firstone, the resultant magnetic force. magnitude N direction ---Select---towards the wireaway from the wireno direction
(b) Find the magnetic torque on the current loop. magnitude N–m direction ---Select---towards the wireaway from the wireno direction magnitude N direction ---Select---towards the wireaway from the wireno direction

Explanation / Answer

(a) the force on the top edge is up, while the force onthe bottom edge is down. These two forces cancel out. . So the net force is only due to the forces on the right andleft edge. The force between two currents is .    F = k L I1 I2 /r         where  k = / 2 = 2 x 10-7 . So in your case:    force on left edge = 2 x10-7 * 0.182 * 7.76 * 4.16 / 0.0450 =   2.611x 10-5 N to the left, because currents inthe same direction attract each other. . and the force on the right edge = 2 x 10-7 *0.182 * 7.76 * 4.16 / 0.2270 = 0.5176 x10-5 N to the right, because oppositecurrents repel. . So the net force on the loop is     2.611- 0.5176 =    2.093 x 10-5 N    to the left . (b) the nettorque on the loop is zero. This is because all the forcesacting on the loop are parallel to the plane of the loop, so noneof the forces exert a torque on the loop. . So in your case:    force on left edge = 2 x10-7 * 0.182 * 7.76 * 4.16 / 0.0450 =   2.611x 10-5 N to the left, because currents inthe same direction attract each other. . and the force on the right edge = 2 x 10-7 *0.182 * 7.76 * 4.16 / 0.2270 = 0.5176 x10-5 N to the right, because oppositecurrents repel. . So the net force on the loop is     2.611- 0.5176 =    2.093 x 10-5 N    to the left . (b) the nettorque on the loop is zero. This is because all the forcesacting on the loop are parallel to the plane of the loop, so noneof the forces exert a torque on the loop.

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

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