Magnetic force on Earth core and Shear force required for rotation of earth core

 Yes — comparing forces per unit area (in N/m², or Pascals) is very meaningful. This allows you to evaluate pressure-like effects from different mechanisms acting on the inner core’s surface. Let's compare two:

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1. Magnetic Coupling Pressure (from earlier)

Estimated from magnetic torque:

F_{\text{magnetic}} \approx 10^4 \, \text{N/m}^2

This comes from  over the boundary layer (~1 km thick).


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2. Gravitational Pressure from Overlying Earth

The inner core is under tremendous pressure from the weight of Earth above. At the inner core boundary:

P_{\text{gravity}} \approx 330 \, \text{GPa} = 3.3 \times 10^{11} \, \text{N/m}^2

This pressure doesn’t rotate the core, but it’s important context — magnetic stress is millions of times smaller than gravitational pressure.


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3. Shear Stress Needed for Rotation (Friction-like drag)

We can estimate how much tangential (rotational) stress per unit area is required to cause your hypothesized jerky motion of the core. From the torque  and surface area :

\text{Average shear stress} = \frac{\tau}{R \cdot A}
\approx \frac{2.28 \times 10^{23}}{(1.22 \times 10^6)(1.87 \times 10^{13})}
\approx 10^4 \, \text{N/m}^2

Which matches the magnetic force per unit area — consistency check passed.


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Conclusion

Magnetic force per unit area: ~10⁴ N/m²

Pressure from Earth’s weight: ~10¹¹ N/m² (but doesn’t cause rotation)

Shear force required for rotation: also ~10⁴ N/m²