Final answer:
The magnitude of the induced current is 150 mA, and the direction is counterclockwise in the scenario where a conducting rod moves through a magnetic field, generated by moving the rod with a constant velocity in a circuit which includes an 8.0 ohms resistor.
Step-by-step explanation:
The scenario described involves electromagnetic induction, where a conducting rod moves through a magnetic field, inducing an electromotive force (EMF) and consequently a current in the circuit. To find the magnitude of the induced current, we use Faraday's law which states that the EMF (ε) induced in the circuit is equal to the rate of change of the magnetic flux (Φ).
ε = -dΦ/dt = -B(dl/dt) = -Bvl
Where:
B is the magnetic field strength.
l is the length of the rod.
v is the speed of the rod.
Inserting the given values: B = 0.40 T, l = 0.25 m, and v = 12 m/s, we get:
ε = -(0.40 T)(12 m/s)(0.25 m) = -1.2 V
The negative sign indicates that the EMF acts to oppose the change in magnetic flux, according to Lenz's Law. The induced current (I) can be calculated using Ohm's law:
I = |ε|/R, where R is the total resistance in the circuit.
Thus, I = |-1.2 V|/8.0 Ω = 0.15 A or 150 mA
The direction of the induced current can be determined by using the right-hand rule, which says that if we point the thumb of the right hand in the direction of the rod's velocity and the fingers in the direction of the magnetic field, the palm gives the direction of the induced EMF. Since the magnetic field is out of the paper and the rod moves to the left, the palm faces downwards, indicating the current is counterclockwise. Therefore, the correct answer is C. 150 mA counterclockwise.