Respuesta :
Answer:
N = 1238
Explanation:
It is given that,
Change in current in the toroidal solenoid, [tex]\dfrac{dI}{dt}=0.05\ A/s[/tex]
Induced emf in the solenoid, [tex]\epsilon=126\ mV=126\times 10^{-3}\ V[/tex]
Current in tire, I = 1.4 A
Flux in the solenoid, [tex]\phi=0.00285\ Wb[/tex]
The induced emf in the solenoid is given by :
[tex]\epsilon=L\dfrac{dI}{dt}[/tex]
L is the self inductance of the solenoid
[tex]L=\dfrac{\epsilon}{dI/dt}[/tex].........(1)
The self inductance of solenoid is given by :
[tex]L=\dfrac{N\phi}{I}[/tex]............(2)
From equation (1) and (2) :
[tex]\dfrac{N\phi}{I}=\dfrac{\epsilon}{dI/dt}[/tex]
[tex]N=\dfrac{\epsilon I}{\phi (dI/dt)}[/tex]
[tex]N=\dfrac{126\times 10^{-3}\times 1.4}{0.00285\times 0.05}[/tex]
N = 1237.89
or
N = 1238
So, the number of turns in the solenoid is 1238. Hence, this is the required solution.
