Question:

A rectangular coil of length 2 cm and width 1.25 cm with 250 turns carries a current of 55 μA and is subjected to a magnetic field of strength 0.64 T. Work done in rotating the coil by 180° against the torque is:

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For a coil in magnetic field, full rotation work depends only on change in potential energy: \(W = 2mB\).
Updated On: Jun 20, 2026
  • 2.2 μJ
  • 3.5 μJ
  • 4.4 μJ
  • 5.5 μJ
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The Correct Option is C

Solution and Explanation

Step 1: Concept of magnetic dipole work.
A current-carrying coil behaves like a magnetic dipole having magnetic moment: \[ m = N I A \] Work done in rotating a dipole from \(0^\circ\) to \(180^\circ\) is: \[ W = 2mB \]

Step 2: Convert area into SI units.

\[ A = 2 \times 1.25 = 2.5 \, \text{cm}^2 = 2.5 \times 10^{-4} \, \text{m}^2 \]

Step 3: Calculate magnetic moment.

\[ m = 250 \times 55 \times 10^{-6} \times 2.5 \times 10^{-4} \]

Step 4: Simplify step-by-step.

First multiply current and area: \[ 55 \times 10^{-6} \times 2.5 \times 10^{-4} = 137.5 \times 10^{-10} = 1.375 \times 10^{-8} \] Now multiply by turns: \[ m = 250 \times 1.375 \times 10^{-8} = 3.4375 \times 10^{-6} \, \text{A m}^2 \]

Step 5: Work done for 180° rotation.

\[ W = 2mB \] \[ W = 2 \times 3.4375 \times 10^{-6} \times 0.64 \] \[ W = 4.4 \times 10^{-6} \, \text{J} \]

Step 6: Final interpretation.

\[ \boxed{4.4 \, \mu \text{J}} \]
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