An air-filled toroidal solenoid has a mean radius of 15.5 cm and a cross-section

Question

An air-filled toroidal solenoid has a mean radius of 15.5 cm and a cross-sectional area of 5.02 cm2 as shown in (Figure 1). Picture this as the toroidal core around which the windings are wrapped to form the toroidal solenoid. The current flowing through it is 11.9 A , and it is desired that the energy stored within the solenoid be at least 0.385 J

1. What is the least number of turns that the winding must have?

Express your answer numerically, as a whole number, to three significant figures.

Explanation / Answer

1) The effective volume of toroid, V = A*2*pi*r

= 5.02*2*pi*15.5

= 489 cm^3

= 4.89*10^-4 m^3

Energy density required, u = total energy/Volume

= 0.385/(4.89*10^-4)

= 787 J/m^3

let B is the required magnetic field.

we know, u = B^2/(2*mue)

B = sqrt(u*2*mue)

= sqrt(787*2*4*pi*10^-7)

= 0.0445 T

magnetic field inside a toroid, B = N*mue*I/(2*pi*r)

==> N = B*2*pi*r/(mue*I)

= 0.0445*2*pi*0.155/(4*pi*10^-7*11.9)

= 2.90*10^3 turns <<<<<<------Answer

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