In the figure below, the current in the long, straight wire is Jj = 5.20 A and t
ID: 1453809 • Letter: I
Question
In the figure below, the current in the long, straight wire is Jj = 5.20 A and the wire lies in the plane of the rectangular loop, which carries a current I2 = 10.0 A. The dimensions in the figure are c = 0.100 m, a - 0.150 m, and = 0.670 m. Find the magnitude and direction of the net force exerted on the loop by the magnetic field created by the wire, magnitude muN The figure below is a cross-sectional view of a coaxial cable. The center conductor is surrounded by a rubber layer, an outer conductor, and another rubber layer. In a particular application, the current in the inner conductor is Ji = 1.06 A out of the page and the current in the outer conductor is I2 = 2.88 A into the page. Assuming the distance d = 1.00 mm, answer the following.Explanation / Answer
The long wire produces a concentric magnetic field. Using the right-hand rule one can determine that the magnetic field vector is perpendicular to the plane of the loop as marked in the figure. From symmetry one can also predict that the force exerted on the two sides of the loop which are perpendicular to the long wire are opposite. The magnetic force exerted on the entire loop result only from the interaction of the sides parallel to the long wire.
Again using right hand rule one can predict that the direction of the force exerted on the side closer to the long wire is attractive and the force exerted on the farther side is repulsive. Since the magnetic field is inversely proportional to the distance between the interacting wires, the force exerted on the closer side of the loop is stronger than the force exerted on the farther side of the loop. Hence the magnetic force exerted on the entire wire is toward the long wire and the magnitude of the magnetic force is
Fb = i2.l*(uoi1 / 2pic)*sin90 - i2.l(uoi1 / 2pi(c + a))sin90
Fb = (uoi1i2l / 2pi)*(1/c - 1/(c+a))
Fb = [(4pi x 10^-7 x 5.2 x 10 x 0.670) / 2pi]*(1/0.100 + 1/(0.100 + 0.150)
Fb = 41.8 x 10^-6 N
Fb = 41.8 uN