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 0.65 V and a current of 3.4 A are induced in the coil. The wire is the 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?
emf 1 V
current 2 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 in a closed loop is equal to the negative rate of change of the magnetic flux through the loop.
For the circular coil:
emf1 = -d(Φ)/dt
Given that emf1 = 0.65 V, we can solve for the rate of change of magnetic flux.
Next, consider the circular coil. Since the magnetic field and its rate of change remain the same, the square coil will experience the same change in magnetic flux as the circular coil. Both coils have the same size and orientation within the magnetic field. Therefore, the emf induced in the square coil will also be 0.65 V.
So, the emf induced in the square coil is 0.65 V (same as the circular coil).
For the current in the square coil, we can use Ohm's law, which states that current (I) is equal to the ratio of emf (E) to resistance (R) in a circuit.
I = E/R
Since the wire is re-formed into a single-turn square coil, the resistance remains the same. Therefore, the current induced in the square coil will also be the same as the circular coil, which is 3.4 A.
Therefore, the current induced in the square coil is 3.4 A (same as the circular coil).
In summary:
emf induced in the square coil = 0.65 V
current induced in the square coil = 3.4 A