Lenz's Law — Question 9

Question 9

A moving rod polarity fixes the current sense and the braking-force direction

A rod slides right at speed v on horizontal rails separated by ell in a field $-B\hat z$. A resistor R closes the circuit at the left. Neglect self-inductance and rail resistance.

1. Find the positive-charge force, rod polarity and current direction through the rod.

2. Find the rod magnetic-force vector and verify its sign.

Question 9 - Solution

A moving rod polarity fixes the current sense and the braking-force direction

(a) Method and calculation

$$\vec v\times\vec B=Bv\hat y\Rightarrow V_{\mathrm{top}}>V_{\mathrm{bottom}}.$$$$\boxed{i=B\ell v/R\quad\text{upward through the rod, CCW around the loop}}.$$

(b) Interpretation and further calculation

$$\vec F=i\ell\hat y\times(-B\hat z)=-i\ell B\hat x.$$$$\boxed{\vec F=-B^2\ell^2v\hat x/R}.$$

Physical check

The charge-separation picture and the flux-change rule give the same counterclockwise current.