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Unformatted text preview: 43. Consider an inﬁnitesimal segment of the loop, of length ds. The magnetic ﬁeld is perpendicular to the
segment, so the magnetic force on it is has magnitude dF = iB ds. The horizontal component of the
force has magnitude dFh = (iB cos θ) ds and points inward toward the center of the loop. The vertical
component has magnitude dFv = (iB sin θ) ds and points upward. Now, we sum the forces on all the
segments of the loop. The horizontal component of the total force vanishes, since each segment of wire
can be paired with another, diametrically opposite, segment. The horizontal components of these forces
are both toward the center of the loop and thus in opposite directions. The vertical component of the
total force is
Fv = iB sin θ ds = (iB sin θ)2πa .
We note the i, B , and θ have the same value for every segment and so can be factored from the integral. ...
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