Complete Question
A parallel plate capacitor creates a uniform electric field of 5 x 10^4 N/C and its plates are separated by 2 x 10^{-3}'m. A proton is placed at rest next to the positive plate and then released and moves toward the negative plate. When the proton arrives at the negative plate, what is its speed?
Answer:
Explanation:
From the question we are told that:
Electric field
Distance
At negative plate
Generally the equation for Velocity is mathematically given by
Therefore
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 = A
where
Now,
(a) To calculate the maximum emf:
We know that maximum emf induced in the coil is given by:
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: