A magnetic field is perpendicular to the plane of a single-turn circular coil. The magnitude of the field is changing, so that an emf of 0.63 V and a current of 2.7 A are induced in the coil. The wire is then re-formed into a single-turn square coil, which is used in the same magnetic field (again perpendicular to the plane of the coil and with a magnitude changing at the same rate). What emf and current are induced in the square coil?
help is greatly appreciated!
ε = 0.63V. I = 2.7 A.
ε = -dÔ/dt =-d(B•S•cosα)/dt.
α= 0 => cos α =1,
ε = - S •dB/dt.
If S is the area of the circle,
then S = π •r², => r=sqrt(S/π).
Perimeter (the length of wire) is
L = 2 •π•r = 2 •π• sqrt(S/π).
the side of the square is
L/4 =2 •π• sqrt(S/π)/4 = π• sqrt(S/π)/2
The area of the square is
S1 = {π• sqrt(S/π)/2}² = π²•S/ 4•π = =π•S/4.
ε1= - S1 •dB/dt= -(π•S/4)• dB/dt,
ε1/ε = [(π•S/4) • (dB/dt)] / {S •dB/dt}= =π/4.
ε1 = ε• π/4 = 0.63• π/4 = 0.495 V.
ε = I•R, => R = ε/I = 0.63/2.7 = 0.23 Ω,
ε1 = I1•R,
I1 = ε1/R = 0.495/0.23 = 2.12 A.
To determine the emf and current induced in the square coil, we can use Faraday's Law of electromagnetic induction. According to Faraday's Law, the induced emf (ε) in a coil is equal to the rate of change of magnetic flux through the coil.
In this scenario, we are given the magnitude of the induced emf (0.63 V) and the current (2.7 A) induced in the initial single-turn circular coil. Therefore, we can start by finding the magnetic flux through the circular coil.
The formula for magnetic flux (Φ) through a coil is given by:
Φ = B * A * cosθ
Where:
B is the magnitude of the magnetic field,
A is the area of the coil, and
θ is the angle between the magnetic field and the perpendicular to the plane of the coil.
In this case, since the magnetic field is perpendicular to the plane of the coil, the angle θ is 0°, and cosθ becomes 1. Therefore, the formula simplifies to:
Φ = B * A
Now, let's find the magnetic field B.
Since the magnitude of the induced emf and the current are both changing at the same rate for the circular coil, we can say that the induced emf and the current are directly proportional. Therefore, we can set up the following equation:
ε1 / ε2 = I1 / I2
Where:
ε1 and ε2 are the induced emfs for the circular and square coils, respectively,
I1 and I2 are the currents for the circular and square coils, respectively.
Substituting the given values into the equation, we have:
0.63 V / ε2 = 2.7 A / I2
Now we can solve for ε2 (the induced emf in the square coil) in terms of the known values (0.63 V and I2):
ε2 = (0.63 V * I2) / 2.7 A
Finally, we can substitute the value of ε2 into the formula for the magnetic flux to find the magnetic field B for the square coil:
Φ2 = B * A
The magnetic field B is the same for both the circular and square coils since they are used in the same magnetic field with the same magnitude changing at the same rate. So, we can equate Φ1 for the circular coil and Φ2 for the square coil:
B1 * A1 = B2 * A2
Given that the number of turns in both the circular and square coils is 1, the areas (A1 and A2) of the coils will be different since their shapes are different. The area of a square coil is equal to the side length squared (A2 = s^2).
Now, we can solve for the current (I2) in the square coil using the magnetic field (B2) and the calculated emf (ε2):
I2 = ε2 / B2
Hence, by using the given induced emf and current in the circular coil, along with the principles of electromagnetic induction, we can determine the induced emf and current in the square coil.