Answer:

Explanation:
For n-=1 state hydrogen energy level is split into three componets in the presence of external magnetic field. The energies are,
,
,

Here, E is the energy in the absence of electric field.
And
are the highest and the lowest energies.
The difference of these energies

is known as Bohr's magneton.
B=2.5 T,
Therefore,

Now,

Therefore, the energy difference between highest and lowest energy levels in presence of magnetic field is 
In scientific notation", that number would be written as
6.81 x 10⁻⁴ .
Answer:
Explanation:
The magnetic field is straight up. It is reducing . As per Lenz's law , direction of induced current is such that it opposes the reason which creates it . magnetic field in upper direction is reducing . So current will be such that magnetic field produced by it increases magnetic field in upper direction . In other words , induced current should create magnetic field in upward direction. It is possible when direction of induced current is anti - clockwise, when seen from above.
Answer:
The peak emf of the generator is 40.94 V.
Explanation:
Given that,
Number of turns in primary coil= 11
Number of turns in secondary coil= 18
Peak voltage = 67 V
We nee to calculate the peak emf
Using relation of number of turns and emf


Where, N₁ = Number of turns in primary coil
N₂ = Number of turns in secondary coil
E₂ = emf across secondary coil
Put the value into the formula


Hence, The peak emf of the generator is 40.94 V.