How Does the Stress Tensor Change at Point P Near Current-Carrying Wires?

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SUMMARY

This discussion focuses on calculating the components of Maxwell's stress tensor at a field point P located on the yz plane, specifically at x=0 and a distance y above the xz plane, due to two parallel current-carrying wires. The stress tensor is defined using the electric field E and magnetic field B components, with the equations provided for B due to each wire. The user seeks clarification on the existence of an electric field generated by the currents and whether the tensor components involving the x-axis are zero at point P.

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  • Understanding of Maxwell's equations and stress tensor formulation
  • Familiarity with electromagnetic fields generated by current-carrying conductors
  • Knowledge of vector calculus and tensor notation
  • Proficiency in using mathematical tools for electromagnetic field calculations
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vwishndaetr
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So I'm working on yet another problem, and have come to a minor stump.

2 wires run parallel with the Z axis in the xz plane, one with current I-1, other with current I-2.

I need to determine the components of Maxwell's stress tensor at a field point P, where P is a point on the yz plane (x=0), a distance y above the xz plane.

We know,

T_{ij} = \epsilon_0 \left( E_i E_j - \frac{1}{2} \delta_{ij} E^2 \right) + \frac{1}{\mu_0} \left( B_i B_j - \frac{1}{2} \delta_{ij} E^2 \right)

I calculate B due to wire 1 to be,

\overrightarrow{B_1} = -\frac{{\mu}_0 I_1y}{2\pi(d^2/4+y^2)}\hat{x} -\frac{{\mu}_0 I_1d}{4\pi(d^2/4+y^2)}\hat{y}

And B due to wire 2,

\overrightarrow{B_2} = -\frac{{\mu}_0 I_2y}{2\pi(d^2/4+y^2)}\hat{x} + \frac{{\mu}_0 I_2d}{4\pi(d^2/4+y^2)}\hat{y}

I know the the equation for the stress tensor involves Electric field also, but is there an E-field created by to wires with current? Or is there no E-field and E=0?

Also, since P lies on yz plane, and x=0, does that mean all tensor components with x in the indices is also zero?

Some help to guide me?

Thanks!
 
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