Answer:
(a) 1.58 V
(b) 0.0126 Wb
(c) 0.0493 V
Solution:
As per the question:
No. of turns in the coil, N = 400 turns
Self Inductance of the coil, L = 7.50 mH = 
Current in the coil, i = ![1680cos[\frac{\pi t}{0.0250}]](https://tex.z-dn.net/?f=1680cos%5B%5Cfrac%7B%5Cpi%20t%7D%7B0.0250%7D%5D)
A
where
Now,
(a) To calculate the maximum emf:
We know that maximum emf induced in the coil is given by:
![e = L\frac{d}{dt}(1680)cos[\frac{\pi t}{0.0250}]](https://tex.z-dn.net/?f=e%20%3D%20L%5Cfrac%7Bd%7D%7Bdt%7D%281680%29cos%5B%5Cfrac%7B%5Cpi%20t%7D%7B0.0250%7D%5D)
![e = - 7.50\times 10^{- 3}\times \frac{\pi}{0.0250}\times \frac{d}{dt}(1680)sin[\frac{\pi t}{0.0250}]](https://tex.z-dn.net/?f=e%20%3D%20-%207.50%5Ctimes%2010%5E%7B-%203%7D%5Ctimes%20%5Cfrac%7B%5Cpi%7D%7B0.0250%7D%5Ctimes%20%5Cfrac%7Bd%7D%7Bdt%7D%281680%29sin%5B%5Cfrac%7B%5Cpi%20t%7D%7B0.0250%7D%5D)
For maximum emf, 
should be maximum, i.e., 1
Now, the magnitude of the maximum emf is given by:
(b) To calculate the maximum average flux,we know that:
(c) To calculate the magnitude of the induced emf at t = 0.0180 s:
Answer:
The kinetic energy of the anti proton is 147.4 MeV.
Explanation:
Given that,
Energy = 2.12 GeV
Kinetic energy = 96.0 MeV
We need to calculate the kinetic energy of the anti proton
Using formula of energy
We know that,
So, 
Put the value into the formula
Hence, The kinetic energy of the anti proton is 147.4 MeV.
Because your hand is warm and the warmness melts the chocolate...
Answer:
along the lead, at right angles to the lead at the middle, and at right angles to the lead at one end.
