A 400 turn solenoid with a length of 20 cm and a radius of 2

A 400 turn solenoid with a length of 20 cm and a radius of 2.0 cm carries a current of 2.0 A. A second coil of four turns is wrapped tightly about this solenoid so that it can be considered to have the same radius as the solenoid. Find the following when the current in the solenoid increases to 5.0 A in a period of 0.90 s. (a) the change in the magnetic flux through the coil

Solution

phi = B A cos theta = mu₀ [N/L] I pi r²

= 4 pi * 10-7 *  [400/0.20] * 2 * pi * 0.02²

= 6.32 * 10^-6 T.m²

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E = - N [delta phi / delta t]

= 4 mu₀ *  [N/L] pi r² [delta I / delta t]

= (4 * 4 pi * 10-7 *  [400/0.20] * pi * 0.02²) * (3 / 0.90)

= 4.21 * 10^-5 V