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A long straight wire carries a current I = 2 A. A semicircular conducting rod rests on two long, frictionless conducting rails (of negligible resistance) that run parallel to the wire. The wire, the rod and the rails all lie in one horizontal plane. The two ends of the semicircular rod are at perpendicular distances of 1 cm and 4 cm from the wire. At t = 0 the rod begins to slide along the rails (parallel to the wire) at a constant speed v = 3.0 m/s. A resistor R = 1.4 ohm in series with an initially uncharged capacitor C0 = 5.0 microF is connected across the rails. Which statement(s) is/are correct? [mu0 = 4*pi*10⁻⁷ SI; ln 2 = 0.7] (A) The maximum current through R is 1.2*10⁻⁶ A. (B) The maximum current through R is 3.8*10⁻⁶ A. (C) The maximum charge on C0 is 8.4*10⁻¹² C. (D) The maximum charge on C0 is 2.4*10⁻¹² C.

1. (A) The maximum current through R is 1.2*10⁻⁶ A.
2. (B) The maximum current through R is 3.8*10⁻⁶ A.
3. (C) The maximum charge on C0 is 8.4*10⁻¹² C.
4. (D) The maximum charge on C0 is 2.4*10⁻¹² C.

Correct answer: (A) The maximum current through R is 1.2*10⁻⁶ A.

Solution

The element of the rod at distance x sees field B = mu0*I/(2*pi*x). Each element moves with speed v parallel to the wire, so the EMF contribution is dE = B*v*dx integrated from 1 cm to 4 cm. EMF = (mu0*I*v/(2*pi)) * ln(4/1) = (mu0*I*v/(2*pi)) * 2*ln2. The maximum current is EMF/R (capacitor acts as a short at t=0), and the maximum charge is C0*EMF (capacitor fully charged, no current).

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