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Solutions for class #3 from Yosumism website
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Problem 9:
Electromagnetism }Current Directions The opposite currents cancel each other, and thus the induction (and field) outside is 0.
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Problem 10:
Electromagnetism }Image Charges The conductor induces image charges -q and -2q since it is grounded at. Since these are (mirror) image charges, each charge induced is the same distance from the conducting plane as its positive component. The net force on q is just the magnitude sum of the positive charge 2q and the two induced charges,
∑ =^ + + = 2
2 2 2 2 2 4
(^4) a q a a a q F πε o πε o Why is it the magnitude sum? Well, recall that , where is the vector pointing from the charge one is interested in to the field charge, i.e., the other charge. If one has two positive charges, then the force points along that unit vector. However, if one has a minus charge and a positive charge, the force points in the other direction. Thus, all the force quantities are additive, and one might as well just take the magnitude sum.)
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Problem 12:
Electromagnetism }Potential A Potential V is related to the electric field E by. Since the problem supplies the approximation tool that the planes are quite large, one can assume the field is approximately constant. The remaining parameter that can't be thrown out by this approximation is the angle, and thus the only choice that yields is choice (B).
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Problem 13:
Electromagnetism }Maxwell's Equations Magnetic monopoles remain a likeable (even lovable) theoretical construct because of their ability to perfectly symmetrize Maxwell's equations. Since the curl term has an electric current, the other curl term should have a magnetic current. ( is taken to be obvious in presence of magnetic charge.) The answer is thus (D), and the revised equations are,
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Problem 18:
Lab Methods }Coax Cable Elimination time. The first- pass question to answer is why is it important that a coax cable be terminated at an end : (A) Perhaps... (B) Probably not. Terminating the cable at an end would not help heat dissipation and thus should not prevent overheating. (C) Perhaps... (D) Probably not, since termination should attenua te the signal rather than to prevent it. (E) Probably not, since image currents should be canceled by the outer sheath. Choices (A) and (C) remain. Now, use the second fact supplied by ETS. The cable
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Problem 34:
Electromagnetism }TEM Waves The full formalism of a conducting cavity can be solved via TEM (transverse electromagnetic) wave guides. However, to solve this problem, one needs only the two boundary conditions from the reflection at a conducting surface, and . The electric field parallel to the cavity is the transverse field, and thus one has choice (D), exactly the conditions above.
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Problem 36:
Electromagnetism }Boundary Conditions The conductor perfectly reflects the incoming wave, and none is transmitted. The electric field is thus reversed. However, since E and B are perpendicular (related to each other by the Poyting Vector where the direction of propagation is given by the direction of ), the magnitude of B is increased by 2, but its direction stays the same. Search on the GRE Physics Solutions homepage with keyword conductors for more on this.
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Problem 54:
Electromagnetism }Faraday Law The induced current would act, according to Lenz Law, to oppose the change. In this case, since the field is decreasing (the wire is being pulled away from the field), the induced current would act to increase the field. On the side closest to the long wire, it would thus point in the same direction as the current from the long-wire. This eliminates all but choices (D) and (E). Now, since the rectangular loop wire can not induce a force on itself, the force is due to the field from the long wire. To the left of the loop, the long wire has a field pointing into the page, and thus the force there is left-wards. One can check again that choice (E) is right by right- hand- ruling the field on the right side of the loop. Since the field due to the long wire is again into the page, the force here is towards the right (since the current runs down the page on the right side of the loop).
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