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Magnetic field due to parallel wires
Two very long thin wires carrying equal and opposite currents +/-I are p laced parallel to the x-axis at y=0 and z=+/-a. Find an expression for B field at a point P in the xy-plane (z=0) and show that its maximum gradient occurs for y = a/sqrt(3).
Field due to long wire at distance d away:
B=(mu_0*I)/(2*PI*d)
By symmetry, i expect B field components to cancel except in y direction..So i end up getting total field:
B=B_y=(mu_0*I)/(PI*d) * cos(theta)
where theta is the angle that d makes to the y axis. Because of geometry of situation, this is equivalent to saying:
B=B_y=(mu_0*I*y)/(PI*d^2)
Then for the gradient to be maximized, I need maximum of:
dB/dtheta=-(mu_0*I)/(PI*d) * sin(theta)
...which should occur when sin(theta) is a max...but i don't understand how i could show that the maximum is for y = a/sqrt(3) any thoughts..? thankyou!
Homework Statement
Two very long thin wires carrying equal and opposite currents +/-I are p laced parallel to the x-axis at y=0 and z=+/-a. Find an expression for B field at a point P in the xy-plane (z=0) and show that its maximum gradient occurs for y = a/sqrt(3).
Homework Equations
Field due to long wire at distance d away:
B=(mu_0*I)/(2*PI*d)
The Attempt at a Solution
By symmetry, i expect B field components to cancel except in y direction..So i end up getting total field:
B=B_y=(mu_0*I)/(PI*d) * cos(theta)
where theta is the angle that d makes to the y axis. Because of geometry of situation, this is equivalent to saying:
B=B_y=(mu_0*I*y)/(PI*d^2)
Then for the gradient to be maximized, I need maximum of:
dB/dtheta=-(mu_0*I)/(PI*d) * sin(theta)
...which should occur when sin(theta) is a max...but i don't understand how i could show that the maximum is for y = a/sqrt(3) any thoughts..? thankyou!
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