To give a solution to this exercise, it is necessary to apply Gauss's magnetic law in which he proposes the magnetic flux through closed surface is zero.
Our data given are as follows:
Inward flux = 13Wb
Outward flux = 22Wb
Since the field of the second flow is directed OUT of the curved side we need to subtract the values, therefore,
Flux = 22-13
Answer:
6.93 volts
Explanation:
q = magnitude of charge on proton = 1.6 x 10⁻¹⁹ C
r = radius of the orbit = 2.08 x 10⁻¹⁰ m
V = Electric potential due to proton on electron
Electric potential due to proton on electron is given as
Inserting the values
V = 6.93 volts