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**1. Homework Statement**

This is a question regarding Problem 7.11 in Griffiths Electromagnetism book, which considers a falling square loop of aluminum through a uniform magnetic field pointing into the page with only part of the loop in the field at t=0 (see picture below).

To solve this, normally one calculates that there is an upward magnetic force of IlB, where I is the current, l is the length of the top portion of the wire, and B is the constant magnetic field. To get I, you'd calculate the emf ξ using the flux rule and then divide by the resistance of the aluminum, R.

However, this is the way to solve this problem using the flux rule to get the emf. I'd like to get the emf by calculating the work done per unit charge (as Griffiths does previously in the chapter).

**2. Homework Equations**

F

F

_{mag}= q(

**v**x

**B**)

**3. The Attempt at a Solution**

As the loop starts to fall, its velocity is (initially) all in the negative

**z-hat**direction so the magnetic force pushes charges to the right, creating a current (

**u**in the diagram). However, consequently, the charges now take on a diagonal path (

**w**). So the magnetic force is then = quB(

**z-hat**) + qvB(

**x-hat**) (blue in the diagram). Now I can get the correct emf if I assume u is constant, because with qu (or really λu) being the current I, the total work done here is IlB (the top part is l and the sides don't contribute).

My question is why can I assume u is constant? If u is the current per electron, doesn't u increase since there is a magnetic force of qvB(

**x-hat**) on a given electron, hence increasing it's speed?

Everywhere I've looked people solve this problem using the flux rule, but I haven't seen it done this way.