Question

At time \(t\), the cardiac dipole is oriented at \(-45^\circ\) (minus forty five degrees) to the horizontal axis. The magnitude of the dipole is \(3\) mV. Assuming Einthoven frontal plane configuration, what is the magnitude (in mV) of the electrical signal in lead II? (Round off the answer to two decimal places.)

Hint:

Lead voltages in the frontal plane are simple projections of the cardiac dipole onto lead axes: \(V = M\cos(\Delta\theta)\). Lead II is aligned at \(+60^\circ\).

Answer 1

Correct Answer: 2.8

Step 1: In the Einthoven triangle, the axis of Lead II is at \(+60^\circ\) with respect to the horizontal (Lead I at \(0^\circ\)). The measured lead voltage equals the projection of the dipole vector on the lead axis: \[ V_{\text{lead}} = M \cos(\theta_{\text{dipole}}-\theta_{\text{lead}}). \] Step 2: Here \(M=3\) mV, \(\theta_{\text{dipole}}=-45^\circ\), \(\theta_{\text{lead}}=60^\circ\). Hence \[ |V_{II}| = 3\,|\cos(60^\circ-(-45^\circ))| = 3\,|\cos(105^\circ)| = 3\times 0.2588 \approx 0.78\ \text{mV}. \] Therefore, the magnitude of the Lead II signal is \(\boxed{0.78\ \text{mV}}\).