Peter has a square shaped coil which include two turns of wires. One day he takes this coil to his physics lab in his school. He wants to find the voltage generated in the coil when he placed the coil inside a varying magnetic field. So he connected a voltmeter to the coil and placed the coil inside a magnetic field that varies 6mT every 10ms. The angle between the plane of the coil and magnetic field is 60 degrees. What is the value of the induced voltage display in the volt meter?
To find the value of the induced voltage in the voltmeter, we can use Faraday's law of electromagnetic induction. According to this law, the induced voltage in a coil is equal to the rate of change of magnetic flux through the coil.
The formula to calculate the induced voltage is:
E = -N * dφ/dt
where:
E is the induced voltage
N is the number of turns in the coil
dφ/dt is the rate of change of magnetic flux
In this case, we are given that the coil has 2 turns and the magnetic field varies by 6 mT (millitesla) every 10 ms (milliseconds). We also know the angle between the plane of the coil and the magnetic field is 60 degrees.
First, we need to calculate the rate of change of magnetic flux. The magnetic flux is given by the formula:
φ = B * A * cos(θ)
where:
B is the magnetic field strength
A is the area of the coil
θ is the angle between the plane of the coil and the magnetic field
Since the coil is square-shaped, the area is equal to the side length squared:
A = s^2
Given that the coil is square and the number of turns is 2, we can assume that both sides of the square coil are parallel to the field lines and use one of the sides as the length of the coil. Therefore, the area can be expressed as:
A = s^2 = (L * W)
where:
L is the length of the coil side
W is the width of the coil side
From the given information, we know that the angle between the plane of the coil and the magnetic field is 60 degrees.
Now, let's calculate the rate of change of magnetic flux:
dφ/dt = dB/dt * A * cos(θ)
The rate of change of the magnetic field is given as 6 mT every 10 ms, which can be converted to 6 * 10^-3 T / 10 * 10^-3 s = 0.6 T/s.
Plugging in the values, we have:
dφ/dt = 0.6 T/s * (L * W) * cos(60)
Since the coil is square-shaped, the length and width are equal, so we can substitute L for W:
dφ/dt = 0.6 T/s * (L^2) * cos(60)
Now we have the rate of change of the magnetic flux. Let's calculate the induced voltage:
E = -N * dφ/dt
Plugging in the values, we have:
E = -2 * 0.6 T/s * (L^2) * cos(60)
At this point, we need to know the value of the length of the coil side (L) to calculate the induced voltage accurately.