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Wireless energy using magnetic inductive coupling. So magnetic field will be affect around device. We use it to charge mobile device so I wonder if it will affect compass in our device

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Theoretically yes, practically no. Practical radiating fields have frequencies above 10kHz, thus the magnetic fields are switching direction at high frequency and so any disturbing torque will also switch direction at this frequency. A compass needle has too much inertia to respond to such high frequency fields.

A simple mathematical model is

$$I\,\mathrm{d}_t^2 \theta +\mu\,B_\oplus\,\theta + \delta\,\mathrm{d}_t\,\theta= \mu\,B_w\,\cos(\omega\,t)$$

Where:

  1. $\theta$ is the needle's angular displacement relative to its equilibrium position (where it is aligned to the Earth's magnetic field);
  2. $\mu$ is the needle's magnetic moment;
  3. $\delta$ models dissipative drag (damping) in the system (often arising from steeping the needle in oil so that its oscillations dampen quickly allowing a quick reading)
  4. $B_\oplus$ is the Earth's magnetic field component along the needle's length
  5. $I$ the mass moment of inertia of the needle about its pivot and
  6. $B_w$ is the amplitude of the wireless magnetic field.

Now solve for $\theta(t)$ and you find that the oscillation amplitude is:

$$\hat{\theta}=\frac{\mu\,B_w}{\sqrt{(\mu\,B_\oplus - I\,\omega^2)^2+\delta^2\,\omega^2}}$$

and thus falls off very swiftly with frequency. The resonant frequency $\omega_0=\sqrt{\frac{\mu\,B_\oplus}{I}}$ and, as you will know from experience (what frequency does a compass needle swing to and fro at?), is a few Hz at the most. So at 10kHz, the amplitude will be roughly eight orders of magnitude smaller than its resonant value.

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