The two are equivalent if the wire is not moving. Using the Leibniz integral rule and that div B = 0, results in, ∮ ∮ ∂ ∂ Σ Σ ( t ) d ℓ ℓ ⋅ ⋅ F / q ( r , t ) = − − ∫ ∫ Σ Σ ( t ) d A ⋅ ⋅ ∂ ∂ ∂ ∂ t B ( r , t ) + ∮ ∮ ∂ ∂ Σ Σ ( t ) v × × B d ℓ ℓ and using the Maxwell Faraday equation, ∮ ∮ ∂ ∂ Σ Σ ( t ) d ℓ ℓ ⋅ ⋅ F / q ( r , t ) = ∮ ∮ ∂ ∂ Σ Σ ( t ) d ℓ ℓ ⋅ ⋅ E ( r , t ) + ∮ ∮ ∂ ∂ Σ Σ ( t ) v × × B ( r , t ) d ℓ ℓ since this is valid for any wire position it implies that, F = q E ( r , t ) + q v × × B ( r , t ) .