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This is problem 7.7c from David J. Griffiths - Introduction to Electrodynamics.

A metal bar of mass $m$ slides frictionlessly on two parallel conducting rails a distance $l$ apart. A resistor $R$ is connected across the rails, and a uniform magnetic field $B$, pointing into the page, fills the entire region.

The force on the bar is $F = \frac{B^2l^2v}{R}$ (to the left).

If the bar starts out with speed $v_0$ (to the right as in the figure) at time $t = 0$, and is left to slide, what is its speed at a later time $t$?

The correct solution is:

$\frac{dv}{dt} = -\frac{B^2l^2v}{Rm} \Rightarrow v = v_0e^{-\frac{B^2l^2t}{Rm}}$

But my initial solution was: $v = v_0 - \frac{B^2l^2vt}{Rm} \Rightarrow v = \frac{v_0}{1+\frac{B^2l^2t}{Rm}}$

I think the formula $v = v_0 + at$ is only valid when acceleration is constant. That is where my error lies. Right?

  • $\begingroup$ Yes, that formula holds only for constant acceleration. $\endgroup$ – G. Smith Jul 27 at 17:56

The formula you're using is only valid when the acceleration of the body is constant.

You can read up more here to understand why this is so. Simply put it's because of our assumption of constant acceleration that we end with the result you're using otherwise you'll get a differential equation.

In this question the force keeps on changing hence the acceleration also keeps changing.

To get the correct answer as mentioned in the book you need to write a differential equation for the velocity.

You can do this by writing your acceleration as


As it is rate of change of velocity You'll end up with a fairly simple differential equation and by taking the antilog on both sides you'll get your answer.


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