Magnetic Properties of Matter: Practice Problem & Solution
6. A 250 turn rectangular coil of length $2.1\text{ cm}$ and width $1.25\text{ cm}$ carries a current of $85\text{ }\mu\text{A}$ and subjected to a magnetic field of strength $0.85\text{ T}$. Work done for rotating the coil by $180^\circ$ against the torque is: (2017-Delhi)
Solution Explained:
To solve this problem, we apply the core principles of Magnetic Properties of Matter. Understanding the underlying formula is key to arriving at the correct answer below:
Magnetic moment $M = NIA = 250 \times (85 \times 10^{-6}\text{ A}) \times (2.1 \times 1.25 \times 10^{-4}\text{ m}^2) \approx 5.58 \times 10^{-6}\text{ A m}^2$.
Work done $W = MB(1 - \cos 180^\circ) = 2MB = 2 \times (5.58 \times 10^{-6}) \times 0.85 \approx 9.1\text{ }\mu\text{J}$.
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