Answer:
the average induced emf in the second winding is [tex]7.9168*10^{-5}V[/tex]
Explanation:
The magnetic field inside the first solenoid is given by,
[tex]B= \mu_0NI[/tex]
Where [tex]\mu[/tex] is the permeability of the free space
N is the number of turns of solenoid per unit length
I is the current in the solenoid
A is the cross-sectional area of the wire
Replacing we have,
[tex]B= (4*\pi*10^{-7})(90/cm (\frac{100cm}{1m}))(0.350A)[/tex]
[tex]B = 3.9584*10^{-3}T[/tex]
Thus average emf induced in the second windigs is,
[tex]\epsilon{avg}=\N\frac{dB}{dt}[/tex]
[tex]\epsilon_{avg}=\frac{d}{dt}(AB)[/tex]
[tex]\epsilon_{avg}=A\frac{dB}{dt}[/tex]
[tex]\epsilon_{avg}= A\frac{dB}{dt}[/tex]
[tex]\epsilon_{avg}=(8*10^{-4})\frac{3.9584*10^{-3}}{0.04}[/tex]
[tex]\epsilon_{avg} = 7.9168*10^{-5}V[/tex]
Therefore the average induced emf in the second winding is [tex]7.9168*10^{-5}V[/tex]
