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For a rectangular loop whose plane is parallel to \[\vec{B}\] (θ = 90°), the torque is computed as force × perpendicular distance between the two forces, which gives:

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Question

For a rectangular loop whose plane is parallel to \[\vec{B}\] (θ = 90°), the torque is computed as force × perpendicular distance between the two forces, which gives:

Options

  • \[\tau = 0\]

  • \[\tau = BIb \times b = BIb^2\]

  • \[\tau = BIb \times l = BIA\]

  • \[\tau = BIl \times l = BIl^2\]

MCQ
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Solution

The two forces of magnitude \[F_1 = BIb\] on sides BC and AD are separated by the length l of the loop. Hence \[\tau = F_1 \times l = BIb \times l = BIA\], since \[A = l \times b\]. For N turns, this becomes \[\tau_{max} = NIAB\].

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