MEC O8

Solution – Question 8

Question 8: Torque on a Current Loop

The loop is released on a frictionless table. The magnetic forces acting on the segments within the field will create a torque about the Center of Mass (CM) of the square loop.

Calculation of Lever Arm and Torque

To determine the rotation, we must calculate the torque exerted by the magnetic forces about the Center of Mass (CM).

1. Location of Pivot (CM):
The square loop has side length $l$. The Center of Mass is located at the geometric center, which is at a horizontal distance of $l/2$ from the left edge.

2. Force on a Segment:
Consider a horizontal segment of length $u$ inside the magnetic field. The magnetic force acts at the center of this segment (since the field is uniform).

  • Center of force position: $u/2$ from the left edge.
  • Magnitude of Force: $F = I u B$.

3. Lever Arm Calculation:
The lever arm ($r$) is the horizontal distance between the pivot (CM) and the point where the force acts.

$$ r = \text{Position of CM} – \text{Position of Force Center} $$ $$ r = \frac{l}{2} – \frac{u}{2} = \frac{l – u}{2} $$

4. Torque Equation:
The torque ($\tau$) contributed by a segment of length $u$ is:

$$ \tau = \text{Force} \times \text{Lever Arm} $$ $$ \tau = (I u B) \times \left( \frac{l – u}{2} \right) $$ $$ \tau = \frac{IB}{2} [ u(l – u) ] $$

Analysis of Rotation

The total torque depends on the function $f(u) = u(l – u)$, which is a downward-opening parabola with a maximum at $u = l/2$.

  • Top Segment ($u = x$): Generates a Clockwise torque (Force Up, left of CM).
  • Bottom Segment ($u = y$): Generates an Anticlockwise torque (Force Down, left of CM).

The net rotation follows the segment with the larger value of $u(l-u)$.

Evaluating Case (a) & (b):
We are given $x < y < l/2$. Since both $x$ and $y$ are less than $l/2$, they are on the "rising" side of the parabola where increasing length increases the function value.

$$ \text{Since } y > x \implies y(l-y) > x(l-x) $$

Therefore, the torque from the bottom segment ($\tau_y$) is stronger than the torque from the top segment ($\tau_x$). The net torque is in the direction of the bottom force: Anticlockwise.

Correct Answers: (a), (b), and (d)
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