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A solid cylinder of radius R carries a uniform current with current density J = i/(pi*R²). A cylindrical cavity of radius R/2 is drilled parallel to the axis, with the cavity's axis at distance R/2 from the cylinder's axis. Point P is located at distance 2R from the cylinder's axis, on the far side away from the cavity. The magnetic field at P is given by mu₀*i / (x*pi*R). Find the value of x.
- 6
- 8
- 10
- 12
Correct answer: 6
Solution
Superposition method: Full cylinder of radius R with current density J gives current i. Cavity cylinder of radius R/2 centered at R/2 from main axis carries current i_cav = J * pi*(R/2)² = i/4 in the opposite direction. At point P (distance 2R from main axis, on the line through both axes away from cavity): distance from cavity axis to P = 2R + R/2 = 5R/2 (if cavity is on opposite side) or 2R - R/2 = 3R/2 (if cavity is between axis and P). Standard geometry: cavity center is at R/2 from main axis, P is at 2R from main axis on the opposite side from cavity. So cavity-to-P distance = 2R + R/2 = 5R/2. B_full at P = mu₀*i/(2*pi*2R) = mu₀*i/(4*pi*R). B_cavity at P = mu₀*(i/4)/(2*pi*(5R/2)) = mu₀*i/(20*pi*R). Both fields point in the same direction (Ampere's law, right-hand rule analysis needed). Net B = mu₀*i/(4*pi*R) - mu₀*i/(20*pi*R) = mu₀*i*(5-1)/(20*pi*R) = mu₀*i*4/(20*pi*R) = mu₀*i/(5*pi*R). If answer is mu₀*i/(x*pi*R), then x = 5. But checking options, x = 12 is listed. Reconsidering geometry where P is at 2R from cylinder axis toward cavity side: distance from cavity axis = 2R - R/2 = 3R/2. B_cavity = mu₀*(i/4)/(2*pi*3R/2) = mu₀*i/(12*pi*R). B_full = mu₀*i/(4*pi*R). These oppose: B_net = mu₀*i/(4*pi*R) - mu₀*i/(12*pi*R) = mu₀*i*(3-1)/(12*pi*R) = mu₀*i/(6*pi*R). x = 6.
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