Three questions on Ampere’s Law

1. The long, straight wire AB shown in Figure (Please see the attached file) carries a current of 14 A. The rectangular loop whose long edges are parallel to the wire carries a current of 5 A. Find the magnitude and direction of the net force exerted on the loop by the magnetic field of the wire.

2. A long straight solid cylinder, oriented with its axis in the z direction, carries a current whose current density is J. The current density, although symmetrical about the cylinder axis, is not constant and varies according to the relation,

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J = (b/r) exp[(r-a)/d] for r < = a and
J = 0 for r >= 0.

Let Io be the total current passing through the entire cross section of the wire. This problem asks to find an expression for Io and to find the magnetic field at various distances from the center of the cylinder using Ampere’s law. Please see the attached JPG file for the rest of this question.

3. A ring-shaped conductor with radius a = 2.5 cm has a total positive charge Q = 0.125 nC uniformly distributed around it, as shown in the diagram (See the attached file). The center of the ring is at the origin of coordinates O.

(a) What is the electric filed (magnitude and direction) at point P, which is on the x-axis at x = 40.0 cm?

(b) A point charge q = -2.50uC is placed at the point P described in part a. What are the magnitude and direction of the force exerted by the charge q on the ring?

See attached file for full problem description.

Solution Preview
Following is the text part of the solution. Please see the attached file for complete solution. Equations, diagrams, graphs and special characters will not appear correctly here.

 

From Ampere’s law B=uo I/2 pi r (here uo = mu zero)

B at ‘a’ is B1= uo I1/2(pi)a
B at ‘b’ is B2= uo I1/2(pi)b

Force between two current carrying wires is attractive if the currents run in same direction. Force is repulsive if the currents run in opposite direction. So in the above diagram, F1 and F2 are marked to reflect this.

Force due to I1 on the left wire is F1=B1*IL = [uo I1/2(pi)a] I2*l
F2 = [uo I1/2(pi)b] I2*l

b>a therefore, F1>F2 therefore, the net force on the rectangular wire is in the direction of F1 (-x direction)
Magnitude of the force is F1-F2

Note that the forces on the shorter sides of the rectangular wire is as marked. The net value of these forces is zero.

Now all the quantities in F1 and F2 are known, the student can substitute these values to find the net …

Solution Summary
Three college level problems on electricity and magnetism have been solved. Well-written answers are in a 4-page word document. Each problem accompanies a diagram.To continue with the answer check on mycoursewriter.com/

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