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Two resistanceless parallel rails are joined at one end by an inductor of inductance L. A magnetic field B fills the region, perpendicular to the plane of the rails. A conducting bar of length l and mass m, lying across the rails, is given an impulse J toward the right. Choose the correct option(s) about the subsequent motion (energy is shared between kinetic energy of the bar and magnetic energy of the inductor).

1. Velocity of the bar is half its initial value after a displacement d = sqrt(3 J² L / (4 B² l² m))
2. Current through the inductor when the bar's velocity is half the initial value is i = sqrt(3 J² / (4 L m))
3. Velocity of the bar is half its initial value after a displacement d = sqrt(3 J² L / (B² l² m))
4. Current through the inductor when the bar's velocity is half the initial value is i = sqrt(3 J² / (m L))

Correct answer: Velocity of the bar is half its initial value after a displacement d = sqrt(3 J² L / (4 B² l² m))

Solution

The bar+inductor loop satisfies flux relation L*i = B*l*x (taking x as displacement), so i = B l x / L. Energy is conserved: (1/2)m v0² = (1/2)m v² + (1/2)L i², with initial speed v0 = J/m. Setting v = v0/2: (1/2)m v0² - (1/2)m (v0/2)² = (1/2)L i² -> (3/8)m v0² = (1/2)L i². Substituting v0 = J/m gives i² = (3/4) J²/(L m), i.e. i = sqrt(3 J²/(4 L m)). Then x from i = B l x / L gives x = L i/(B l), and squaring/substituting yields d = sqrt(3 J² L/(4 B² l² m)). So the displacement option with the factor 4 in the denominator is correct (and the matching current is sqrt(3J²/(4Lm))).

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