Two Experiments — Question 10

Question 10

Instrument time averaging can hide opposite induction pulses

A detector reports average emf over a window W. A pickup produces +E for time tau, then -E for time tau, and zero during the remainder, with $W\ge2\tau$. A faster detector can resolve each pulse.

1. Find the average and root-mean-square emf over W.

2. For E=.400 V, tau=2.00 ms and W=20.0 ms, evaluate both and the heat in a 10.0-ohm resistive pickup.

Question 10 - Solution

Instrument time averaging can hide opposite induction pulses

(a) Method and calculation

$$\langle\mathcal E\rangle=(E\tau-E\tau)/W=\boxed0.$$$$\mathcal E_{\mathrm{rms}}=\sqrt{(E^2\tau+E^2\tau)/W}=E\sqrt{2\tau/W}.$$

(b) Interpretation and further calculation

$$\boxed{\mathcal E_{\mathrm{rms}}=.400\sqrt{.2}=.178885\,\mathrm V}.$$$$H=\frac{\mathcal E_{\mathrm{rms}}^2W}{R}=\boxed{6.40\times10^{-5}\,\mathrm J}.$$

Physical check

A zero average detector output is not sufficient evidence that induction was absent.