I'm trying to understand the following example which is taken from "Introduction to Electrodynamics" by David J. Griffiths:enter image description hereenter image description here

Problem: The calculation of emf is clear. We take a snapshot of the loop and determine $$\text{emf} = \oint \vec{f}.d\vec{l}$$where $\vec{f}$ is the force per unit charge. I don't understand how the work done per unit charge is calculated. The vector $d\vec{l}$ points straight up, so why do we have the following equation?$$\int \vec{f}_{pull}.d\vec{l}= (uB)(\frac{h}{\cos \theta})\sin \theta$$ How do we know that $\vec{u}$(vertical velocity) is constant? This calculation is really confusing to me and it seems there are lot of unstated assumptions.


He assumes that the circuit has a resistance, so the current due to the Lorentz force is constant. In the case of charges in free space they would be accelerated.

Once accepting that their velocity in the wire is $u$, the elementary work done by who is pushing the loop is $dE = \mathbf {F.dx} = (\mathbf u \times \mathbf B) \mathbf {.dx}$

$\mathbf u \times \mathbf B$ has a magnitude of $uB$ because the vectors are perpendicular and is horizontal. The displacement of charges $\mathbf {dx}$ has the direction of the vectorial sum of the velocity of the loop $v$ and the velocity of the charges in the wire $u$. If the resultant $w$ has an angle $\theta$ with the vertical, the elementary displacement is $\frac{dl}{cos{\theta}}$. And the angle of the horizontal force with $w$ is $sin(\theta)$.

As the magnitudes and angles are constants, it is an integral of $dl$ resulting in $uB\frac{h}{cos(\theta)}sin(\theta)$


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