A rod of length 5 cm moves perpendicular to a uniform magnetic field of 2×10⁻⁴ Wb m⁻². If the acceleration of the rod is 2 m s⁻², the rate of increase of the induced emf is .

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(4) 20×10⁻⁶ V s⁻¹. Given: ℓ = 5 cm = 0.05 m; B = 2×10⁻⁴ Wb m⁻²; acceleration a = 2 m s⁻². Idea: the motional emf is ε = Bℓ v. With B and ℓ fixed, it grows only because v grows. (dε)/(dt) = Bℓ (dv)/(dt) = Bℓ a. (dε)/(dt) = 2×10⁻⁴ × 0.05 × 2 = 2×10⁻⁵ = 20×10⁻⁶ V s⁻¹. The unit follows from the working: volts per second. So the emf increases at 20×10⁻⁶ V s⁻¹.

A rod of length 5 cm moves perpendicular to a uniform magnetic field of 2×10⁻⁴ Wb m⁻². If the acceleration of the rod is 2 m s⁻², the rate of increase of the induced emf is ______.

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