Question

At time t, the cardiac dipole is oriented at - 45° (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:

Always remember the standard angles for Einthoven's leads (I: 0°, II: 60°, III: 120°). The measured voltage is always the dipole magnitude multiplied by the cosine of the angle *between* the dipole vector and the lead axis.

Answer 1

Step 1: Concept:
An ECG lead records the projection of the cardiac dipole vector along the axis of that lead. The recorded voltage is given by: \[ V = |\vec{M}| \cos(\phi) \] where $\phi$ is the angle between the dipole direction and the lead axis.
Step 2: Given Data:

• Dipole magnitude $|\vec{M}| = 3 \text{ mV}$
• Dipole angle $\theta_d = -45^\circ$
• Lead II angle $\theta_{L2} = +60^\circ$

Step 3: Angle Difference: \[ \phi = \theta_{L2} - \theta_d = 60^\circ - (-45^\circ) = 105^\circ \]
Step 4: Voltage in Lead II: \[ V_{II} = 3 \cos(105^\circ) \] Using $\cos(105^\circ) \approx -0.2588$, \[ V_{II} = 3 \times (-0.2588) = -0.7764 \text{ mV} \] Since the magnitude is required, \[ |V_{II}| \approx 0.78 \text{ mV} \] Final Answer: The magnitude of the signal in Lead II is $0.78 \text{ mV}$.