A 2.10 mH toroidal solenoid has an average radius of 5.80 cm and a cross-sectional area of 2.20 cm^2 .A.How many coils does it have? In calculating the flux, assume that B is uniform across a cross section, neglect the variation of B with distance from the toroidal axisB.At what rate must the current through it change so that a potential difference of 2.40 V is developed across its ends?

Answer :

A toroidal solenoid has 1664 turns.

The rate of current is across it is 1142.86 A/s.

What is a solenoid?

A solenoid is a machine made of a casing, a coil of wire, and a moving plunger (armature). A magnetic field that generates around the coil when an electrical current is applied pulls the plunger in. A solenoid, to put it simply, transforms electrical energy into mechanical work.

Given,

Length L = 2[tex]\pi[/tex]r

= 2[tex]\pi[/tex]×0.058 = 0.3644 m

L = [tex]u_{o}[/tex][tex]N^{2}[/tex]A/l

N= [tex]\sqrt{LXl/u_{oA} }[/tex]

N = [tex]\sqrt{2.1X10^{-3}X0.3644/4\pi X10^{-7}X2.2X10^{-4}}[/tex]

N= 1664 turns

B) V = L [tex]\frac{dl}{dt}[/tex]

[tex]\frac{dl}{dt}[/tex] = [tex]\frac{V}{L}[/tex] = [tex]\frac{2.4}{2.1X10^{-3} }[/tex]

[tex]\frac{dl}{dt}[/tex] = 1142.86 A/s

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