A long, straight wire carries a current of 740 mA. A thin metal rod 45 cm long is oriented perpendicular to the wire and moves with a speed of 1.8 m/s in a direction parallel to the wire. What are the size and direction of the emf induced in the rod if the nearest point of the rod is 3.5 cm away from the wire, and if the rod moves in a direction parallel to the current?
To find the magnitude of the electromotive force (emf) induced in the rod, we can use Faraday's law of electromagnetic induction. According to Faraday's law, the emf induced in a conductor is equal to the rate of change of magnetic flux through the surface bounded by the conductor.
First, we need to find the magnetic field generated by the current-carrying wire at the position of the rod. We can use the Biot-Savart law to calculate the magnetic field produced by a current-carrying wire. The formula to calculate the magnetic field at a distance 'r' from a long, straight wire is given by:
B = (μ₀ * I) / (2π * r)
Where:
B is the magnetic field
μ₀ is the permeability of free space (4π x 10^-7 Tm/A)
I is the current flowing through the wire
r is the distance from the wire
Given:
I = 740 mA = 0.74 A (convert milliamperes to amperes)
r = 3.5 cm = 0.035 m (convert centimeters to meters)
Plugging the values into the formula, we have:
B = (4π x 10^-7 Tm/A * 0.74 A) / (2π * 0.035 m)
B = (0.74 * 10^-7 Tm) / (0.07 m)
B ≈ 1.057 x 10^-5 T
The magnetic field at the position of the rod is approximately 1.057 x 10^-5 Tesla.
Now, to find the emf induced in the rod, we multiply the magnetic field by the length of the rod and the velocity of the rod. However, we need to take into account the angle between the velocity of the rod and the magnetic field.
Since the rod moves parallel to the wire carrying current, the angle between the velocity and the magnetic field is 90 degrees.
The formula to calculate the emf induced in the rod is given by:
emf = B * l * v * sin(θ)
Where:
emf is the electromotive force
B is the magnetic field
l is the length of the rod
v is the velocity of the rod
θ is the angle between the velocity and magnetic field
Given:
B = 1.057 x 10^-5 T
l = 45 cm = 0.45 m (convert centimeters to meters)
v = 1.8 m/s
θ = 90 degrees
Plugging the values into the formula, we have:
emf = (1.057 x 10^-5 T) * (0.45 m) * (1.8 m/s) * sin(90 degrees)
emf = (1.057 x 10^-5 T) * (0.45 m) * (1.8 m/s) * 1
emf ≈ 8.593 x 10^-6 V
The emf induced in the rod is approximately 8.593 x 10^-6 volts.
Regarding the direction of the induced emf, it follows Lenz's law. Lenz's law states that the direction of the induced current (and therefore the induced emf) is such that it opposes the change that produced it. In this case, the induced emf would create a current that creates a magnetic field opposing the magnetic field produced by the current in the wire.