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For this problem on the topic of magnetic fields, we are told that a copper rod with a mass of 1 kilogram is resting on two horizontal rails that are separated by a meter, and the rod carries a current of 50 amperors from one rail to the other.
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The coefficient of static friction between the rod and rails is 0 .6, and we want to find the magnitude and the angle of the smallest magnetic field that puts the rod on the verge of sliding.
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Now, the magnetic force must push horizontally in the rod to overcome the force of friction, but can be oriented so that it also pulls up on the rod.
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And thereby reduces both the normal force and the force of friction.
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The forces acting on the rod are f, the force of the magnetic field, mg, the magnitude of the force of gravity, fn the normal force exerted by the stationary rails upward on the rod, and little f, which is the horizontal force of friction.
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Now, we'll assume that the rod is on the verge of moving eastward, which means that little f points westward, and therefore capital f has an eastward component fx and an upward component fy, which can be related to.
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To the components of the magnetic field once we assume a direction for the current in the rod.
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And so again, we assume the current flows northward, and then by the right -hand rule, a downward component bd of the mini -ary field b will produce the eastward force fx, and the west wind component bw will produce the upward f -y...