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Unformatted text preview: Solutions for Hand-In Set #3 PHYS 212 - Spring 2007 Problem A15 — Mapping the field of a permanent magnet (a) Sketch of a bar magnet with direction of magnetic field around the magnet: (b) Real magnet with compass in various locations nearby: Overall, direction of compass agreed fairly well with my predictions. A few of the angles were a bit off. And from the compass directions, it appears as though the S side of the magnet is on the left here. Problem A18 — Rail guns The work-enegy helps out here: W = Δ K. In this case the force on the bar is F = IHB (to the left), and this constant force is applied over a distance L , so the work energy theorem gives IHBL = 1 2 mv 2 f- , or v f = r 2 IHBL m . Comment: The solution given above makes it appear that magnetic forces are actually doing work here, but that can’t be right! The result derived above is actuallly correct, but the analysis is simplistic. We leave it as a challenge to you to figure out what’s really going on here. Problem A19 — Cancelling the Earth’s magnetic field North Side view of coil Current out of page Current into page The magnetic field on the axis of a single current loop is given by B = μ 4 π 2 πR 2 I ( x 2 + R 2 ) 3 / 2 ....
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- Spring '08
- Magnetic Field, 2m, 0.3 m, 3 m, 3-cm, 1.48 k