Answer:
the average induced emf in the second winding is ![7.9168*10^{-5}V](https://tex.z-dn.net/?f=7.9168%2A10%5E%7B-5%7DV)
Explanation:
The magnetic field inside the first solenoid is given by,
![B= \mu_0NI](https://tex.z-dn.net/?f=B%3D%20%5Cmu_0NI)
Where
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,
![B= (4*\pi*10^{-7})(90/cm (\frac{100cm}{1m}))(0.350A)](https://tex.z-dn.net/?f=B%3D%20%284%2A%5Cpi%2A10%5E%7B-7%7D%29%2890%2Fcm%20%28%5Cfrac%7B100cm%7D%7B1m%7D%29%29%280.350A%29)
![B = 3.9584*10^{-3}T](https://tex.z-dn.net/?f=B%20%3D%203.9584%2A10%5E%7B-3%7DT)
Thus average emf induced in the second windigs is,
![\epsilon{avg}=\N\frac{dB}{dt}](https://tex.z-dn.net/?f=%5Cepsilon%7Bavg%7D%3D%5CN%5Cfrac%7BdB%7D%7Bdt%7D)
![\epsilon_{avg}=\frac{d}{dt}(AB)](https://tex.z-dn.net/?f=%5Cepsilon_%7Bavg%7D%3D%5Cfrac%7Bd%7D%7Bdt%7D%28AB%29)
![\epsilon_{avg}=A\frac{dB}{dt}](https://tex.z-dn.net/?f=%5Cepsilon_%7Bavg%7D%3DA%5Cfrac%7BdB%7D%7Bdt%7D)
![\epsilon_{avg}= A\frac{dB}{dt}](https://tex.z-dn.net/?f=%5Cepsilon_%7Bavg%7D%3D%20A%5Cfrac%7BdB%7D%7Bdt%7D)
![\epsilon_{avg}=(8*10^{-4})\frac{3.9584*10^{-3}}{0.04}](https://tex.z-dn.net/?f=%5Cepsilon_%7Bavg%7D%3D%288%2A10%5E%7B-4%7D%29%5Cfrac%7B3.9584%2A10%5E%7B-3%7D%7D%7B0.04%7D)
![\epsilon_{avg} = 7.9168*10^{-5}V](https://tex.z-dn.net/?f=%5Cepsilon_%7Bavg%7D%20%3D%207.9168%2A10%5E%7B-5%7DV)
Therefore the average induced emf in the second winding is ![7.9168*10^{-5}V](https://tex.z-dn.net/?f=7.9168%2A10%5E%7B-5%7DV)
Eh? i don’t understand huhu
Answer:5
Explanation:
Given
First Plate has a charge of +Q and area A
Second Plate has a charge of -3 Q and area A
We Know electric Field due to sheet charge is given by
![E=\frac{Q}{2A\epsilon }](https://tex.z-dn.net/?f=E%3D%5Cfrac%7BQ%7D%7B2A%5Cepsilon%20%7D)
Where Q=charge over the Plate
A=Area of plate
= Permittivity of free space
Electric Field Due to Positive charge will always be away from it while for negative charge it is towards it.
Net Electric Field at a point between between the Plates is the superimposition of two electric with direction
![E_{net}=\frac{Q}{2A\epsilon }+\frac{3Q}{2A\epsilon }](https://tex.z-dn.net/?f=E_%7Bnet%7D%3D%5Cfrac%7BQ%7D%7B2A%5Cepsilon%20%7D%2B%5Cfrac%7B3Q%7D%7B2A%5Cepsilon%20%7D)
![E_{net}=\frac{2Q}{A\epsilon }](https://tex.z-dn.net/?f=E_%7Bnet%7D%3D%5Cfrac%7B2Q%7D%7BA%5Cepsilon%20%7D)
Net electric Field is towards the negative charged plate
Answer:
Magnetic field, B = 1.9232 T
Explanation:
Given data:
Value of the charge, Q = 0.026 C
Speed, V = 443.592 m/s
Force experienced, F = 22.182
Now,
the Force (F) experienced by a charge in a magnetic field is given as:
F = QVBsinθ
where,
B is the magnetic field
Angle between the magnetic field and the velocity.
since, the velocity is in horizontal direction and the magnetic field is downwards. Therefore, the angle θ = 90°
thus, we have
22.182 = 0.026 × 443.592 × B × sin90°
or
B = 1.9232 T
Answer:
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