A square loop (side L) spins with angular frequency ω in field of strength B. It is hooked to a load R.
Write an expression for current I(t) in terms of B,L,R,ω and t (enter omega for ω).
(b) How much work is done by the generator per revolution? Express your answer in terms of B,L,R and ω (enter omega for ω).
(c) To make it twice as hard to turn (twice as much work), what factor would you have to multiply the resistance R?
No one has answered this question yet.
a) L^2*B*omega*sin(omega*t)/R
the magnetic flux will be:
Φ = A*B cos(w*t)
Then the electromotrix force will be:
-dΦ/dt = A*B*w*sin(w*t) = ε(t)
Since the current is given by
I(t) = ε(t)/R
we have
I(t) = L^2*B*w*sin(w*t)/R
c)the factor is 1/2
part b plsssss
To find the expression for the current I(t), let's use Faraday's law of electromagnetic induction:
ε = -dΦ/dt
where ε is the induced electromotive force (EMF), Φ is the magnetic flux through the loop, and t is time.
The magnetic flux through a square loop can be expressed as:
Φ = B * A = B * L^2
where B is the magnetic field strength and A is the area of the loop (which is equal to the side of the square loop squared).
Since the loop is spinning with an angular frequency ω, the time derivative of the flux can be written as:
dΦ/dt = d(B*L^2)/dt = 2B*L*dL/dt
The derivative of the side length L with respect to time can be expressed as the derivative of the angle θ with respect to time (dθ/dt) multiplied by the radius of the circle, which is L/2. Therefore:
dL/dt = (L/2) * dθ/dt
Now, we need to substitute these derivatives into the equation for the induced EMF:
ε = -dΦ/dt = -2B*L * (L/2) * dθ/dt = -B*L^2 * dθ/dt
The EMF is equal to the product of the current I(t) and the resistance R:
ε = I(t) * R
Therefore, we can equate both expressions for the EMF to find the expression for the current I(t):
I(t) * R = -B*L^2 * dθ/dt
Rearranging the equation, we get:
I(t) = (-B*L^2 * dθ/dt) / R
So, the expression for current I(t) in terms of B, L, R, ω (angular frequency), and t is:
I(t) = (-B*L^2 * dθ/dt) / R
Moving on to part (b), to find the amount of work done by the generator per revolution, we need to calculate the total energy transferred (or work) during one complete rotation of the loop.
The work done by the generator is given by the equation:
W = ∫(ε * I(t)) dt (from t = 0 to t = T)
where T is the time taken for one complete revolution (period).
Substituting the expression for the EMF, we have:
W = ∫((-B*L^2 * dθ/dt) * I(t)) dt (from t = 0 to t = T)
Since the current I(t) is constant during one revolution (since it is not specified otherwise), we can bring it outside the integral:
W = I(t) * ∫(-B*L^2 * dθ/dt) dt (from t = 0 to t = T)
The integral on the right-hand side can be simplified as:
W = I(t) * (-B*L^2 * ∫dθ) (from t = 0 to t = T)
Since the integral of dθ is just θ, we have:
W = I(t) * (-B*L^2 * [θ] from t = 0 to t = T)
Therefore, the work done per revolution can be expressed as:
W = I(t) * (-B*L^2 * (θ(T) - θ(0)))
Now, moving on to part (c), to make it twice as hard to turn (twice as much work), we need to adjust the resistance R.
Since the work done is directly proportional to the resistance R, to make it twice as hard to turn, we need to double the resistance.
Therefore, the factor by which we have to multiply the resistance R is 2.