Question

A toroidal solenoid is connected to an ac source that has voltage amplitude 12 V and frequency 1 kHz. It has cross-sectional area 1 cm 2, mean radius 10 cm, and resistance 5 Ohm. Ignore the variation of the magnetic field across the cross section of the solenoid. If the amplitude of the current is 1A, what would be the number of turns (N) in the solenoid? (L = Mon?A / (21r) and Mo = 2 x 10-71 · m/A) O 3700 1853 7413 926 463 AAA a much nomer does 6V-battery supply?

Answer 1

Sol: Impedence Z = V/I =12/1 = 12 ohm

L= HON2A :] =3.18 10-11 N 2ar

And $X_{L}=2\pi fL =2\pi \times 1\times 10^{3}\times 3.18\times 10^{-11}N^{2}=2\times 10^{-7}N^{2}$

Now

$Z=\sqrt{R^{2}+X_{L}^{2}}$

or

z? = R' + X2

or

$X_{L}^{2}=Z^{2}-R^{2}=12^{2}-5^{2}=119$

or

$X_{L}=10.91$

But

$X_{L}=2\times 10^{-7}N^{2}$

Hence

$N^{2}=10.91/(2\times 10^{-7})=5.455\times 10^{7}$

or

$N=\sqrt{5.455\times 10^{7}}=7386turn$

Hence third option has closest value.