Referring to the diagram, a long, thin wire carries a current I parallel to the z-axis â, = (0, 0, 1) (the direction of the current is assumed pointing out of the paper). The wire's axis is at the point with position vector ro = (xo, yo, 0),relative to the origin O. The point P shown has position vector r = (x, y, 0), so that the position vector of P relative to the wire's axis is R. Based on the Ampère's law, together with the vectors shown in the figure, we can know that the Cartesian components of the magnetic field H at Pare H_{x}=-\frac{I}{2 \pi} \frac{y-y_{0}}{\left(x-x_{0}\right)^{2}+\left(y-y_{0}\right)^{2}}, \quad H_{y}=\frac{I}{2 \pi} \frac{x-x_{0}}{\left(x-x_{0}\right)^{2}+\left(y-y_{0}\right)^{2}}, \quad H_{z}=0 Assume tuv w are the last four digits of your student number. For example, if your student number isC1700123, then t = 0, u =1. v =2 and w= 3. A square boundary has its corners at the position vectors (10,10,0) m, (10,–10,0) m,(-10,–10,0)m and (-10,10,0) m. Two long, thin wires are fixed at (t+0.5,u +0.5,0) and(-v+0.5, -w+ 0.5,0) within it, with their axes parallel to the z-axis (0, 0, 1). The current /along each wire is 1 A, with the direction being pointing out of the paper. Using the equations given in the question to calculate the magnetic field H at the origin O. Calculate the force per unit length F = I× B on each of the two wires (considering B= Ho H,where uo = 4r ×10 H/m); comment on your result in the context of Newton's third law. Calculate the closed line integral q, H • de on the square boundary.

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