Two Experiments — Question 2

Question 2

Doubling the passage speed changes pulse height but not signed transferred charge

A moving magnet produces pickup linkage $\Lambda(x)$ between fixed positions x1 and x2. One run uses speed v and another uses speed 2v along the same path. For this calibrated experiment, use signed flux linkage $\Lambda=N\Phi$, $\mathcal E=-d\Lambda/dt$ and $i=\mathcal E/R$; neglect pickup self-inductance.

1. Compare the emf at the same position and the passage duration.

2. Compare the signed charge transported between the two endpoint positions.

Question 2 - Solution

Doubling the passage speed changes pulse height but not signed transferred charge

(a) Method and calculation

$$\mathcal E=-\frac{d\Lambda}{dx}v\Rightarrow\boxed{\mathcal E_{2v}(x)=2\mathcal E_v(x)}.$$$$\Delta t=(x_2-x_1)/v\Rightarrow\Delta t_{2v}=\Delta t_v/2.$$

(b) Interpretation and further calculation

$$Q=\int i\,dt=-\frac1R\int d\Lambda=-\frac{\Lambda(x_2)-\Lambda(x_1)}R.$$$$\boxed{Q_{2v}=Q_v}.$$

Physical check

A faster pulse is taller and shorter. The invariance concerns signed charge between the same linkage endpoints, not dissipated heat.