Question

A 2.10 mHmH toroidal solenoid has an average radius of 5.60 cmcm and a cross-sectional area of 2.80 cm2cm2.

Part A

How many coils does it have? In calculating the flux, assume that BB is uniform across a cross section, neglect the variation of BB with distance from the toroidal axis.

Part B

At what rate must the current through it change so that a potential difference of 2.80 VV is developed across its ends?

Express your answer in amperes per second.

Answer 1

given

L = 2.1 mH

= 2.1 x 10^-3 H

radius r = 5.6 cm

= 0.056 m

area A = 2.8 cm²

= 0.00028 m²

a )

using equation L = _o N² A / 2 r

N = ( 2 r L / _o A )^1/2

= ( 2 x 3.14 x 0.056 x 2.1 x 10^-3 / 4 x 3.14 x 10^-7 x 0.00028 )^1/2   

= ( 2 x 0.056 x 2.1 x 10^-3 / 4 x 10^-7 x 0.00028 )^1/2   

= ( 0.056 x 2.1 x 10^-3 / 2 x 10^-7 x 0.00028 )^1/2   

= ( 0.056 x 2.1 / 2 x 10^-4 x 0.00028 )^1/2   

N = 1449.13

means 1449 turns it has

b )

| di/dt | = / L

= 2.8 / 2.1 x 10^-3

= 1333.333 A/sec