Question:
A wire of length '(L)' carries a current '(I)'. If the wire is turned into a square coil of single turn, the maximum magnitude of the torque in a given magnetic field ((\vec{B})) is
A wire of length '(L)' carries a current '(I)'. If the wire is turned into a square coil of single turn, the maximum magnitude of the torque in a given magnetic field ((\vec{B})) is
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For a fixed length, a circular coil provides the maximum area and thus the maximum torque compared to a square coil.
Updated On: Apr 30, 2026
- (\frac{IBL}{16})
- (\frac{IBL}{8})
- (\frac{IBL^2}{8})
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The Correct Option is D
Solution and Explanation
Step 1: Coil Dimensions
Perimeter of square $= 4s = L \implies s = L/4$.
Step 2: Area Calculation
$A = s^2 = (L/4)^2 = L^2/16$.
Step 3: Torque Formula
$\tau_{max} = BIA$.
$\tau = BI(L^2/16) = \frac{IBL^2}{16}$.
Final Answer: (D)
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