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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. - Physics

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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.

संख्यात्मक

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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पाठ 4: Electromagnetic Induction And Alternating Current - Evaluation [पृष्ठ २६१]

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सामाचीर कलवी Physics - Volume 1 and 2 [English] Class 12 TN Board
पाठ 4 Electromagnetic Induction And Alternating Current
Evaluation | Q IV. 1. | पृष्ठ २६१

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