Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problems 19.61, 19.71
Two long straight wires carry currents, both outwards perpendicular to the plane of this page. The left wire (located at x = 0) carries twice as much current as the right wire (located at x = +0.20 m). The magnitude of the total magnetic field at x = +0.30 m is 8.0×10–7 T. Determine (a) the direction of the total magnetic field at x = +0.30 m, and (b) the amount of current in the left wire (which has the greater amount of current). Show your work and explain your reasoning using the properties of magnetic fields, and superposition.
Solution and grading rubric:
- p:
Correct. At x = +0.30 m, the magnetic fields of each wire point upwards, so the total magnetic field there must point upwards. Then adds magnitudes of the magnetic field due to each wire at x = +0.30 m, with the constraint that I1 = 2·I2, and finds that I1 = 0.48 A. - r:
Nearly correct, but includes minor math errors. - t:
Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. Typically sets up the difference in magnitudes of the magnetic field due to each wire at x = +0.30 m. - v:
Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. Typically has only one wire's contribution to the total magnetic field. - x:
Implementation of ideas, but credit given for effort rather than merit. - y:
Irrelevant discussion/effectively blank. - z:
Blank.
Sections 30882, 30883
Exam code: finalEL7a
p: 9 students
r: 3 students
t: 5 students
v: 7 students
x: 5 students
y: 5 students
z: 1 student
A sample "p" response (from student 3420):
A sample "y" response (from student 0003), already looking forward to summer break:
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