A rectangular loop of N turns and of area A is rotated at a frequency F in a uniform magnetic field B. Show that the induced emf is giving by E=Emax, where omega=2πf ad E max =NABW
To show that the induced emf is given by E = Emax, we can use Faraday's law of electromagnetic induction.
Faraday's law states that the induced electromotive force (emf) in a closed loop is equal to the rate of change of magnetic flux through the loop. Mathematically, this can be expressed as:
E = -d(Φ)/dt
Where E is the induced emf, Φ is the magnetic flux through the loop, and t is time.
In this case, the area of the rectangular loop is A, the number of turns is N, and the magnetic field is B. The magnetic flux through the loop can be calculated as the product of the magnetic field and the area of the loop, multiplied by the number of turns:
Φ = BAN
Taking the derivative of Φ with respect to time, we have:
d(Φ)/dt = BAN(dN/dt)
Since the frequency F is defined as the rate of change of the number of turns with respect to time, we can write:
(dN/dt) = 2πF
Substituting this into the derivative of Φ, we get:
d(Φ)/dt = BAN(2πF)
Finally, substituting this into Faraday's law, we obtain:
E = -BAN(2πF)
The negative sign indicates that the induced emf opposes the change in magnetic field, as given by Lenz's law.
Now, we can simplify the expression by introducing the angular frequency ω, which is defined as:
ω = 2πF
Substituting this into the equation for E, we have:
E = -BANω
Since the negative sign indicates the direction of the induced emf, we can ignore it for the magnitude of the emf. Therefore, the induced emf is:
E = BANω
Now, we can substitute the value of ω:
E = BAN(2πF)
And we obtain:
E = Emax
Where Emax = BAN(2πF) = NABω
So, the induced emf is indeed given by E = Emax, where Emax = NABω.