A 500-turn circular coil with an area of 0.050 0 m2 is mounted on a rotating frame that turns at a rate of 20.0 rad/s in the presence of a 0.050 0-T uniform magnetic field that is perpendicular to the axis of rotation. What is the instantaneous emf in the coil at the moment that the normal to its plane is at a 30.0 angle to the field? answer
ℰ=-dΦ/dt=-d(NBAcosωt)/dt=
=NBAωsinωt =
=500•0.05•0.05•20•sin30° =12.5 V
To calculate the instantaneous emf in the coil, we can use the equation for the induced emf in a rotating coil in a magnetic field:
ε = NABωsinθ,
where
ε is the induced emf,
N is the number of turns in the coil (N = 500),
A is the area of the coil (A = 0.0500 m^2),
B is the magnetic field strength (B = 0.0500 T),
ω is the angular velocity of rotation (ω = 20.0 rad/s), and
θ is the angle between the normal to the coil's plane and the magnetic field (θ = 30.0 degrees).
To begin, we need to convert the angle θ to radians:
θ_rad = θ * (π/180) = 30.0 * π/180 = 0.5236 radians.
Now, we can substitute the given values into the formula for ε and solve for the instantaneous emf:
ε = NABωsinθ = (500)(0.0500 m^2)(0.0500 T)(20.0 rad/s)sin(0.5236)
By calculating the expression ε, we can find the value of the instantaneous emf in the coil at the given moment.
To find the instantaneous EMF in the coil, we can use Faraday's Law of electromagnetic induction, which states that the induced EMF in a coil is equal to the rate of change of magnetic flux through the coil.
The formula for the induced EMF is given by:
EMF = -N * d(Φ)/dt
Where:
EMF is the induced electromotive force (volts),
N is the number of turns in the coil,
d(Φ) is the change in magnetic flux through the coil (we'll calculate this),
and dt is the change in time (which is instantaneous, so it's equal to zero).
Given data:
Number of turns, N = 500
Area of the coil, A = 0.0500 m^2
Magnetic field strength, B = 0.0500 T
Rotation rate, ω = 20.0 rad/s
Angle between the coil's normal and the magnetic field, θ = 30.0°
To find the change in magnetic flux, we can use the formula:
d(Φ) = B * dA * cos(θ)
Where dA is the differential area vector, perpendicular to the plane of the coil.
To find dA, we multiply the area of the coil by the cosine of the angle between the coil's normal and the magnetic field:
dA = A * cos(θ)
Now we can substitute the known values into the formula:
dA = 0.0500 m^2 * cos(30.0°)
dA = 0.0500 m^2 * √(3)/2
dA = 0.0433 m^2
Now we can substitute the calculated value of dA back into the formula for d(Φ):
d(Φ) = 0.0500 T * 0.0433 m^2
d(Φ) = 0.00216 T·m^2
Now, we plug in all the values into the formula for the induced EMF:
EMF = -N * d(Φ)/dt
EMF = -500 * 0.00216 T·m^2 / 0
EMF = 0 volts
Therefore, the instantaneous EMF in the coil at the moment when the normal to its plane is at a 30.0° angle to the field is 0 volts.