Complete question:
What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field 0.425 T. (This is 60 rev/s.)
Answer:
The peak emf generated by the coil is 15.721 kV
Explanation:
Given;
Radius of coil, r = 0.250 m
Number of turns, N = 500-turn
time of revolution, t = 4.17 ms = 4.17 x 10⁻³ s
magnetic field strength, B = 0.425 T
Induced peak emf = NABω
where;
A is the area of the coil
A = πr²
ω is angular velocity
ω = π/2t = (π) /(2 x 4.17 x 10⁻³) = 376.738 rad/s = 60 rev/s
Induced peak emf = NABω
= 500 x (π x 0.25²) x 0.425 x 376.738
= 15721.16 V
= 15.721 kV
Therefore, the peak emf generated by the coil is 15.721 kV
Step up transformer is a device which is used to step up the voltage which is input with some value.
This is based upon the principle of mutual inductance and in this the voltage input and voltage output is different because of number of turns.
Here if ideal transformer is given then power input and power output of the transformer must be same as there is no power loss in ideal transformer.
So we can write

here
= 73.37 A
= 4623 V
= 105033 A
now using above equation we will have

solving above we will have

The derived unit for voltage is named volt.
For the cement bag we can say as per its force diagram we will have

here we will have


now we will have

now plug in all data


so the pulling force will be 295 N