Induced voltage in sliding bar

The bar moves through a magnetic field, so a motional emf is induced:

V_emf = ∮ (v × B)·dl

Here v = 10 a_y m/s, B = 3 a_z mWb/m² = 3×10⁻³ a_z T, and the bar lies along the x-direction between the rails at x = 0 and x = 10 cm = 0.1 m, so dl = dx a_x.

v × B = (10 a_y) × (3×10⁻³ a_z) = 3×10⁻² (a_y × a_z) = 0.03 a_x  V/m

V_emf = ∫(x=0→0.1) 0.03 a_x · dx a_x = 0.03 × 0.1 = 3×10⁻³ V

Check with Faraday's law: the loop area increases at rate l v = 0.1 × 10 = 1 m²/s, so |emf| = B (dA/dt) = 3×10⁻³ × 1 = 3 mV.

Answer: V_emf = 3 mV, with the emf directed along +a_x in the bar.