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A wire is bent into an Archimedean spiral described by r(theta) = 1 + (2/pi)*theta for 0 <= theta <= pi, where r is in arbitrary length units and theta is in radians measured from the x-axis. The origin is point P and current I flows through the wire. If B is the magnetic field at P, find the value of (mu₀ * I * ln(3)) / B.
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Correct answer: 8
Solution
For a current element at polar position (r, theta) relative to origin P, the Biot-Savart contribution is dB = (mu₀*I)/(4*pi) * (r*d(theta)/r²) = (mu₀*I)/(4*pi) * d(theta)/r (only the tangential part contributes to B at origin, all pointing in same direction by right-hand rule). B = (mu₀*I)/(4*pi) * integral from 0 to pi of d(theta)/(1+2*theta/pi). Let u = 1+2*theta/pi, du = (2/pi)*d(theta), d(theta) = (pi/2)*du. Limits: u=1 to u=3. B = (mu₀*I)/(4*pi) * (pi/2) * ln(3) = (mu₀*I*ln(3))/8. Therefore (mu₀*I*ln(3))/B = 8.
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