Practice portal › Electromagnetic Induction › Mutual Inductance
Asked in JEE Main 31st Jan 1st Shift 2024 · Standard coil geometries
M=(2√2μ₀l²)/(π L), and with L=l² this reduces to M=(2√2μ₀)/π.
Putting μ₀=4π×10⁻⁷ gives M=8√2×10⁻⁷ H, so x=8√2≈11.3; the printed value 128 corresponds to x², so check this one against the key before use.
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