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A 100-turn circular coil of radius 10 cm carries a current of 3.2 A. Its magnetic moment is approximately:
A10 A m²
B1.0 A m²
C32 A m²
D100 A m²
Answer & Solution
Correct answer: A. 10 A m²
1. Magnetic moment: $m = NIA = NI\pi r^2$.
2. Area: $\pi r^2 = 3.14\times(0.1)^2 = 3.14\times10^{-2}$ m².
3. Substitute: $m = 100\times3.2\times3.14\times10^{-2}$.
4. $m = 100\times3.2\times0.0314 \approx 10$ A m².
_Source: NCERT Class 12 Physics Ch 4 "Moving Charges and Magnetism", p.157_
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