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प्रश्न
A square coil of side 30 cm with 500 turns is kept in a uniform magnetic field of 0.4 T. The plane of the coil is inclined at an angle of 30° to the field. Calculate the magnetic flux through the coil.
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उत्तर
Square coil of side (a) = 30 cm = 30 × 10-2m
Area of square coil (A) = a2 = (30 × 10-2)2 = 9 × 10-2 m2
Number of turns (N) = 500
Magnetic field (B) = 0.4 T
Angular between the field and coil (θ) = 90 – 30 = 60°
Magnetic flux (Φ) = NBA cos 0
= 500 × 0.4 × 9 × 10-2 × cos 60° = 18 × 12
Φ = 9 W b
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