Question

Exercise 30.8 - Enhanced - with Feedback When the current in a toroidal solenoid is changing at a rate of 0.0200 A/s, the magnitude of the induced emfis 12.3 mV. When the current equals 1.50 A, the average flux through each turn of the solenoid is 0.00448 Wb. Part A How many turns does the solenoid have? ANSWER: N =

Answer 1

Magnetic field due to a toroid inside it is $B = \mu_{o}ni$ [where n = no of turns per unit length]

Let the radius of each turn be R and total no of turns is N.

Given:

i = 1.5A

$\phi_{\text{one turn}}=0.00448 Wb$

$\frac{di}{dt}=0.02 A/s$

induced e.m.f =E = 12.3 mV = 0.0123 V

The flux through each turn is ------(i)

one turn = H,ni* #R

Total flux =

o = Moni* R * N

e.m.f induced = $E=\left |\frac{d\phi}{dt} \right |=\mu_{o}n*\pi R^{2}*N*\frac{di}{dt}$ -----(ii)

Dividing the equations (i) and (ii):

$\frac{E}{\phi_{\text{one turn}}}=\frac{\mu_{o}n*\pi R^{2}*N*\frac{di}{dt}}{\mu_{o}ni*\pi R^{2}}$

$\Rightarrow \frac{E}{\phi_{\text{one turn}}}=\frac{N*\frac{di}{dt}}{i}$

$\Rightarrow N= \frac{E*i}{\phi_{\text{one turn}}*\frac{di}{dt}}= \frac{0.0123*1.5}{0.00448*0.02}=205.9$

As the last turn is not complete, so we will take number of turns as 205. [answer]