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Three very long and parallel wires are positioned parallel to the z-axis and carry currents as shown in the figure below with I1 = 5.0A, I2 = 4.0A and I3 = 2.0A. I1 and I3 are into the page while I2 is out of the page.
a) Determine the magnitude and direction of the net magnetic field at point ï¿½Pï¿½.
b) Determine the magnitude and direction of the net force acting on wire 1.
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Added Fri, 07 Aug '15
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The magnetic field lines around a long wire which carries an electric current form concentric circles around the wire. The direction of the magnetic field is perpendicular to the wire and is in the direction the fingers of your right hand would curl if you wrapped them around the wire with your thumb in the direction of the current.

And magnitude is given by the formula

$B =\frac{\mu _{o}I}{2\pi r}$

a) So , the direction of magnetic field by $I_{2}$ and $I_{3}$ at point P will be in x direction whereas the direction of magnetic field by $I_{1}$Â Â at point P will be in -ve x direction

$\vec{B}_{net}=\vec{B}_{1}+\vec{B}_{2}+\vec{B}_{3}$

Â Â $=\frac{\mu _{o}}{2\pi }[\frac{5}{0.5}(-\vec{i})+\frac{4}{0.2}(\vec{i})+\frac{2}{0.2}(\vec{i})]$

Â Â $= 4 \mu T$$\vec{i}$

b) The direction of magnetic field by $I_{2}$Â Â at point 1 will be in -ve x direction whereas the direction of magnetic field by $I_{3}$Â Â at point 1 will be in x direction

$\vec{B}_{net}=\vec{B}_{2}+\vec{B}_{3}$

Â Â $=\frac{\mu _{o}}{2\pi }[\frac{4}{0.3}(-\vec{i})+\frac{2}{0.7}(\vec{i})]$

$= -2.10 \mu T\vec{i}$

Force on wire 1 per unit length $= I_{1}\vec{B_{net}}$

$= 5 X - 2.10 = - 10.5 X 10^{-6} \frac{N}{m}\vec{i}$

Edit
Added Fri, 07 Aug '15
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