The rotating loop in an AC generator is a square 11.0 cm on a side. It is rotated at 60.0 Hz in a uniform field of 0.800 T. Calculate the following quantities as functions of time t, where t is in seconds.

(a) the flux through the loop
mT·m2

(b) the emf induced in the loop
V

(c) the current induced in the loop for a loop resistance of 2.00 Ω
A

(d) the power delivered to the loop
W

(e) the torque that must be exerted to rotate the loop
mN·m

a) ABcos(wt)=(.11*.11)(.8)cos(2*pi*60t)

=.00968cos(377t)
b) ABwsin(wt)=(.11*.11)(.8)(2*pi*60)sin(2*pi*60t)=3.65sin(377t)
c) (ABw/R)sin(wt)=((.11*.11)(.8)(2*pi*60)/2)sin(2*pi*60t)=1.82sin(377t)
d)((ABw)^2/R)(sin(wt))^2=(((.11*.11)(.8)(2*pi*60))^2/2)(sin(2*pi*60t))^2=6.66sin(377t)
e)(A^2B^2w/R)(sin(wt))^2=(((.11*.11)(.8))^2*(2*pi*60)/2)(sin(2*pi*60t))^2=17.6sin(377t)

fdf

To calculate the quantities as functions of time t for a rotating loop in an AC generator, we will use the following formulas:

(a) The flux through the loop:
The flux is given by the formula:
Φ = B * A * cos(ωt)

where
Φ is the flux (in Weber, Wb),
B is the magnetic field strength (in Tesla, T),
A is the area of the loop (in square meters, m^2),
ω is the angular frequency (in radians per second, rad/s), and
t is the time (in seconds, s).

In this case, the loop is a square with a side length of 11.0 cm = 0.11 m.
So, the area of the loop A = (0.11 m)^2 = 0.0121 m^2.

Given:
B = 0.800 T (uniform magnetic field)
ω = 2πf = 2π * 60.0 rad/s = 120π rad/s (angular frequency)

Therefore, the flux through the loop as a function of time t is:
Φ(t) = (0.800 T) * (0.0121 m^2) * cos(120πt) = 0.00968 cos(120πt) Wb

(b) The emf induced in the loop:
The emf (electromotive force) induced in the loop is given by Faraday's Law of electromagnetic induction:
ε = -dΦ(t)/dt

where
ε is the emf induced (in Volts, V),
dΦ(t)/dt is the rate of change of flux with respect to time.

Differentiating Φ(t) with respect to time, we get:
dΦ(t)/dt = -0.00968 * (-120π) * sin(120πt) = 1.1616π sin(120πt) Wb/s

Therefore, the emf induced in the loop as a function of time t is:
ε(t) = -1.1616π sin(120πt) V

(c) The current induced in the loop for a loop resistance of 2.00 Ω:
The current induced in the loop can be calculated using Ohm's Law:
I = ε/R

where
I is the current induced (in Amperes, A),
ε is the emf induced (in Volts, V), and
R is the loop resistance (in Ohms, Ω).

Given:
R = 2.00 Ω (loop resistance)

Therefore, the current induced in the loop as a function of time t is:
I(t) = (-1.1616π sin(120πt)) / 2.00 A

(d) The power delivered to the loop:
The power delivered to the loop is given by the formula:
P = I^2 * R

where
P is the power delivered (in Watts, W).

Therefore, the power delivered to the loop as a function of time t is:
P(t) = (I(t))^2 * R = [(-1.1616π sin(120πt)) / 2.00]^2 * 2.00 W

(e) The torque that must be exerted to rotate the loop:
The torque required to rotate the loop is given by the formula:
τ = ε * I

where
τ is the torque (in Newton meters, Nm).

Therefore, the torque required to rotate the loop as a function of time t is:
τ(t) = (-1.1616π sin(120πt)) * (-1.1616π sin(120πt)) Nm

To calculate the quantities as functions of time, we need to use the relevant formulas and understand the underlying concepts. Let's go through each quantity step by step:

(a) The flux through the loop:
The flux, denoted by Φ (phi), is given by the product of the magnetic field (B) and the area (A) of the loop:
Φ = B * A

Substituting the given values into the equation:
B = 0.800 T (the uniform magnetic field)
A = (11.0 cm)^2 (the area of the square loop)

To calculate the value of the flux as a function of time, we need to recognize that the loop is rotating at a frequency of 60.0 Hz. We can represent the angle of rotation as ωt, where ω is the angular frequency and t is the time:

Angular frequency, ω = 2πf
Substituting the given frequency into the equation:
ω = 2π * 60.0 Hz

Therefore, the flux as a function of time will be:
Φ(t) = B * A * cos(ωt)

(b) The emf induced in the loop:
The induced emf, denoted by ε (epsilon), is given by the rate of change of flux with respect to time:
ε = -d(Φ)/dt

To find the derivative, we take the derivative of the equation for Φ(t) with respect to time ('t'):

d(Φ(t))/dt = -B * A * ω * sin(ωt)

Therefore, the emf as a function of time will be:
ε(t) = -B * A * ω * sin(ωt)

(c) The current induced in the loop for a loop resistance of 2.00 Ω:
Using Ohm's law, we can calculate the current (I) using the equation:
I = ε / R

Substituting the values we have:
I(t) = ε(t) / R
I(t) = (-B * A * ω * sin(ωt)) / R

Plug in the given resistance value (R = 2.00 Ω) to get the current as a function of time.

(d) The power delivered to the loop:
The power (P) delivered to the loop is given by the product of the current (I) and the voltage (V):
P = I * V

Since we don't have the voltage explicitly given, we need to find it using the relationship between emf, resistance, and current:
V = I * R

Plugging this equation into the power equation, we have:
P(t) = I(t) * [I(t) * R]

(e) The torque that must be exerted to rotate the loop:
The torque (τ) required to rotate the loop against the magnetic field is given by the product of the magnetic moment (μ) and the magnetic field (B):
τ = μ * B

The magnetic moment is given by the product of the current (I) and the area (A) of the loop:
μ = I * A

Substituting this into the torque equation, we have:
τ(t) = I(t) * A * B

Again, plug in the values as functions of time to get the torque.