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When learning about the Lornetz force:

$$F = IL \times B$$

Or the motional emf:

$$ V = vBL$$

The texts I read would state that the field is uniform, what if the magnetic field wasn't uniform?

Say, the source was a permanent magnet, or a single wire(increasing distance away from the wire would lead to a weaker field from a closer point).

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You simply would need to integrate. The Lorentz force on a volume current distribution is $$\mathbf{F} = \int dV \; \mathbf{J}\times\mathbf{B}$$ integrated over whole volume of distribution. For surface or line currents, replace $\mathbf{J} \;dV$ by $\mathbf{K} \;da$ or $I dl$.

For motional emf, the expression is $$ e = \int \mathbf{f}_{mag}\cdot d\mathbf{l}$$ around the loop. Here $\mathbf{f}_{mag}$ is magnetic force per unit charge, acting on the infinitesimal element dl, $\mathbf{v\times B}$. However, there is a tricky point here. This $\mathbf{v}$ is total velocity of charge. That means, if the charge flows with respect to wire at velocity $\mathbf{u}$ and the wire moves with velocity $\mathbf{w}$ in the lab, then $\mathbf{v=w+u} $

Edit: Surface currents are distributions of charge flowing over a surface. The current density K may be a function of point on surface. Line currents are just your typical textbook wires having no features of volume. Ideally, like a thread.

The motional emf is easily mistaken to be just $\int\mathbf{w\times B}\cdot d\mathbf{l}$. Thats why I pointed it out.

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  • $\begingroup$ Your point about $\textbf{v=w+u}$ , is that really necessary now? I mean finally when you take the line integral after taking cross product, only $\textbf{w}$ term would contribute $\endgroup$ Apr 22, 2018 at 5:11

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