A rectangular coil of 70 turns, dimensions 0.100 m by 0.200 m and total resistance 10.0 ohms, rotates with angular speed 32.0 rad/s about the y axis in a region where a 1.00 T magnetic field is directed along the x axis. The rotation is initiated so that the plane of the coil is perpendicular to the direction of vector B at t = 0.
(a) Calculate the maximum induced emf in the coil.
_______V
(b) Calculate the maximum rate of change of magnetic flux through the coil.
_______Wb/s
(c) Calculate the induced emf at t = 0.0500 s.
_______V
(d) Calculate the torque exerted on the coil by the magnetic field at the instant when the emf is a maximum.
_______N·m
your answers are wrong
(a) To calculate the maximum induced emf in the coil, we can use Faraday's law of electromagnetic induction:
emf = N * dΦ/dt
Where emf is the induced electromotive force, N is the number of turns in the coil, and dΦ/dt is the rate of change of magnetic flux through the coil.
The magnetic field is directed along the x-axis, and the coil is rotating about the y-axis. Therefore, the magnetic flux through the coil is given by:
Φ = B * A * cosθ
Where B is the magnetic field strength, A is the area of the coil, and θ is the angle between the magnetic field and the normal to the coil.
In this case, the magnetic field strength (B) is 1.00 T, the area of the coil (A) is (0.100 m * 0.200 m), and at t = 0, the angle θ is 90 degrees. So, cosθ = cos(90°) = 0.
Therefore, the maximum induced emf in the coil at t = 0 is:
emf = 70 * 0 * (d/dt)(1.00 T) = 0 V
So, the maximum induced emf in the coil is 0 V.
(b) The rate of change of magnetic flux through the coil can be calculated as:
dΦ/dt = B * dA/dt * cosθ
Since the area A of the coil remains constant, dA/dt = 0. Therefore, the rate of change of magnetic flux is also 0 Wb/s.
So, the maximum rate of change of magnetic flux through the coil is 0 Wb/s.
(c) The induced emf at t = 0.0500 s can be calculated using the same equation as in part (a):
emf = N * dΦ/dt
The rate of change of magnetic flux (dΦ/dt) is still 0 Wb/s, so the induced emf at t = 0.0500 s is also 0 V.
So, the induced emf at t = 0.0500 s is 0 V.
(d) The torque exerted on the coil by the magnetic field can be calculated using the following formula:
τ = N * B * A * sinθ
Where τ is the torque, N is the number of turns, B is the magnetic field strength, A is the area of the coil, and θ is the angle between the magnetic field and the normal to the coil.
At the instant when the emf is a maximum, the angle θ is 0 degrees since the plane of the coil is perpendicular to the magnetic field. Therefore, sinθ = sin(0°) = 0.
Thus, the torque exerted on the coil at the instant when the emf is a maximum is:
τ = 70 * 1.00 T * (0.100 m * 0.200 m) * 0 = 0 N·m
So, the torque exerted on the coil by the magnetic field at the instant when the emf is a maximum is 0 N·m.
To answer this question, we can use Faraday's Law of electromagnetic induction. Faraday's Law states that the induced electromotive force (emf) in a conducting loop is equal to the rate of change of magnetic flux through the loop.
(a) To calculate the maximum induced emf in the coil, we can use the equation:
emf = N * d(flux) / dt
where N is the number of turns in the coil, d(flux) / dt is the rate of change of the magnetic flux through the coil.
The magnetic flux (Φ) through a surface is given by the equation:
Φ = B * A
where B is the magnetic field strength and A is the area of the surface.
In this case, the magnetic field is 1.00 T along the x-axis, and the area of the coil is given by the product of its dimensions (0.100 m * 0.200 m). So,
A = 0.100 m * 0.200 m = 0.020 m^2
Now, let's calculate the rate of change of magnetic flux (dΦ / dt) based on the given rotational condition. The angle (θ) between the magnetic field and the plane of the coil changes with time according to its angular speed (ω) and time (t) as:
θ = ω * t
Therefore, the rate of change of magnetic flux is:
d(Φ)/dt = B * d(A*cos(θ))/dt
Since the area A is constant,
d(Φ)/dt = B * d(cos(θ))/dt = B * sin(θ) * d(θ)/dt
At t=0, the plane of the coil is perpendicular to the magnetic field, so θ = 90°. In this position, sin(90°) = 1.
Now, we can substitute the given values into the equations:
emf = N * d(Φ)/dt = N * B * sin(θ) * d(θ)/dt = 70 * 1.00 T * sin(θ) * (32.0 rad/s)
The maximum value of sin(θ) is 1, so the maximum induced emf is:
(emf)max = 70 * 1.00 T * 1 * (32.0 rad/s) = 2240 V
Therefore, the maximum induced emf in the coil is 2240 V.
(b) The maximum rate of change of magnetic flux through the coil is equal to the derivative of Φ with respect to time:
(dΦ/dt)max = B * d(cos(θ))/dt = B * d(cos(ωt))/dt
Differentiating cos(ωt) with respect to time, we get:
(d(cos(ωt))/dt) = -ω * sin(ωt)
At t = 0, sin(ωt) = sin(0) = 0.
Therefore, the maximum rate of change of magnetic flux is:
(dΦ/dt)max = B * -ω * sin(0) = 0
So, the maximum rate of change of magnetic flux through the coil is 0 Wb/s.
(c) To calculate the induced emf at t = 0.0500 s, we need to calculate the value of sin(θ) at that time.
θ = ω * t = 32.0 rad/s * 0.0500 s = 1.60 rad
sin(θ) = sin(1.60 rad) ≈ 0.999
Substituting the values into the formula:
emf = N * B * sin(θ) * d(θ)/dt = 70 * 1.00 T * 0.999 * (32.0 rad/s)
The induced emf at t = 0.0500 s is:
emf = 70 * 1.00 T * 0.999 * (32.0 rad/s) = 2236 V
Therefore, the induced emf at t = 0.0500 s is 2236 V.
(d) The torque exerted on the coil by the magnetic field at the instant when the emf is at a maximum can be calculated using the equation:
τ = N * A * B * sin(θ) * d(θ)/dt
Since A and B are constants, the torque can be simplified to:
τ = N * A * B * sin(θ) * d(θ)/dt = N * A * B * sin(θ) * ω
At the instant when the emf is at a maximum, sin(θ) = 1 and ω = 32.0 rad/s.
Therefore, the torque exerted on the coil at the instant when the emf is a maximum is:
τ = 70 * 0.020 m^2 * 1.00 T * 1 * 32.0 rad/s = 44.8 N·m
Therefore, the torque exerted on the coil by the magnetic field at the instant when the emf is a maximum is 44.8 N·m.
a. induced current i = emf/R
emf e = NawB coswt
= 70*0.1*0.2 * 32 *1 * cos 32
= 38 volts
b. rate of change of flux = emf = 38 volts
c. emf e= AdB/dt= 72*0.1*0.2 * 1/5
= 0.288 volts
d. torque T = NiABw coswt