The figure shows a rod of length L 11.4 cm that is forced to move at constant sp

Question

The figure shows a rod of length L 11.4 cm that is forced to move at constant speed v 5.93 m/s along horizontal rails. The rod, rails, and connecting strip at the right form a conducting loop. The rod has resistance 0.517 ?; the rest of the loop has negligible resistance. A current 112 A through the long straight wire at distance a 14.5 mm from the loop sets up a (nonuniform) magnetic field throughout loop. Find the (a) magnitude of the emf in volts and (b) current in amperes induced in the loop. (c) At what rate is thermal energy generated in the rod? (d) What is the magnitude of the force that must be applied to the rod to make it move at constant speed? (e) At what rate does this force do work on the rod? ????? (c) Number (d) Number (e) Number Units Units Units

Explanation / Answer

Note that in circumstances such as these, we typically neglect the self-inductance due to the induced current in the wire loop, as its effect is negligible considered to the applied field.

a) We must integrate to determine the total magnetic flux through the loop -- the formula ? = BLv is only applicable when B is uniform.

Let x be the instantaneous displacement of the rod from the strip. Break up the loop into small strips of width dr and length x, each a distance r from the wire. Since these strips are thin, we may assume that B is practically constant on each strip, and equal in magnitude to ??I/(2?r). Hence, each strip contributes a flux d? = B dA = B (x dr) = ??Ix/(2?r) dr. Integration gives:

d? = ??Ix/(2?r) dr
? = ?[r?, r?] ??Ix/(2?r) dr
= ??Ix/(2?) ln(r?/r?)

By Faraday's law:
? = -??/?t = ??I/(2?) ln (r?/r?) dx/dt
= ??Iv/(2?) ln (r?/r?)

At this point, substitute v = 5.93 m/s, r? = 1.45 cm and r? = 12.85 cm to obtain ? = 14.49 x 10^(-4) V [1].

(b) All resistances except the 0.517 ohm are negligible: hence, 14.49 x 10^(-4) V = I(0.517 ohm) and I = 7.49 x 10^(-4) A.

(c) P = I²R = (7.49 x 10^(-4) A)²(0.517 ohm) = 2.9 x 10^(-7) W. (V²/R is equally good, as is P = VI).

(d) Since we assume that self-inductance is negligible, we neglect the magnetic force between the strip and the rod as being much less than the magnetic force from the applied current. Let I? = 112 A be the applied current, and I = 7.49 x 10^(-4) A from part (b). Take i to be to the right, j up and k out of the page to form a right-handed coordinate system. Also notice that the induced current in the loop is clockwise, per Lenz's law.

The magnetic field B due to I? is out of the page at every point below the wire, and has magnitude ??I?/(2?r). Since the current is clockwise, the current in the wire is upward, and the current element I dr points in the +j direction. We have:

dF = I dr × B
= I dr j × [??I?/(2?r) k]
= ??II?/(2?) dr/r i

Integrating this between the r? and r? from before yields a total magnetic force on the rod of ??II?/(2?) ln(r?/r?) = 1.83 x 10^(-8) i N. This force opposes the leftward motion of the rod, so a force of -1.83 x 10^(-8) i N must be applied.

(e) You're correct in noting this is the same as (c). In fact, we may either check this by computing P = Fv (since F and v are in the same direction) = (1.83 x 10^(-8) N)(5.93 m/s) = 1.08 x 10^(-7) W, or indeed, we can USE this equality to determine the applied force must be (1.08 x 10^(-7) W)/(5.93 m/s) = 1.83 x 10^(-8) N, without appealing to the formula dF = I ds × B

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