Unformatted text preview: he right, they are deflected down, and the bottom plate acquires a positive
I I = qE ~ E = vB ::} V = Et = vEt, with the bottom at higher potential.
(c) If negative charges flow to the left, they are also deflected down, and the bottom plate acquires a negative
charge. The potential difference is still the same, but this time the top plate is at the higher potential.
(b) qvB I I (END) 5.40
Problem From Eq. 5.17, F = I J(dl x B). But B is constant, in this case, so it comes outside the integral: F = I (J dl) x B, and J dl = w, the vector displacement from the point at which the wire first enters the field to
the point where it leaves. Since wand B are perpendicular, F = I Bw, and F is perpendicular to w.
The angular momentum acquired by the particle as it moves out from the center to the edge is L= / ~~ dt = / N dt = / (r x F) dt = / r x q(v x B) dt =q / r x (dl x B) = q [/ (r. B) dl- / B(r. dl)] . But r i~ perpendicular to B, so r. B = 0, and r. dl = r. dr = !d(r. r) = !d(r2) = rdr = (lj27r)(27rrdr). SoL = - 2: foR B 27rr = - 2: / B da. It followsthat L = - -!; <I>,
where =J <I>
B da is the total flux.
In particular, if <I> 0, then L = 0, and the charge emerges with zero angular momentum, which means it is
goingalong a radial line.
I I Problem 5.42
From Eq. 5.24, F = J (K x Bave) da. Here K = av, v = wR sin B J;, da = R2 sin B dB d4J, and Bave= !(Bin + Bout). From Eq. 5.68,...
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- Fall '12
- Trigraph, Electric charge, dl