A solenoid 4 cm in diameter and 20 cm in length has 250 turns and carries a current of 15A. Calculate the flux through the surface of a disc of 10 cm radius that is positioned perpendicular to and centered on the axis of the solenoid

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  • 3
B= 4pi ×10^-7 × I x n _______________ L So B= 4 x 3.14×10^-7×15×250/0.2 = 2.35 x 10^-1 Flux (phi) = BAcos theta. Theta= 0 So, Flux = 2.35 x 10^-1 × 3.14 x 4 x 10^-4 = 2.96 x 10^-5 Wb
  • 59
R = radius of solenoid = 2 cm L = 20 cm,    turns = N = 250 turns per unit length = n = 1,250 turns/meter current i = 15 A Magnetic field through the solenoid = μ₀ n i = 4π * 10⁻⁷ * 1,250 * 15  units          B = 2.356 * 10⁻² H The magnetic field outside the turns of the solenoid is zero, ie., it exists only in the area of π r². Flux = B * πr² = 0.0185  Volt-sec
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