A 50-cm wire placed in an East-West direction is moved horizontally to the North with a speed of 2 m/s. The Horizontal and Vertical components of the earth's magnetic field at that location are 25 ìT and 50ìT respectively. What is the emf between the ends of the wire?
A) 30 ìV
B) 50 ìV
C) 20 ìV
D) 40 ìV
E) 10 ìV
30
50
To find the emf between the ends of the wire, we can use Faraday's law of electromagnetic induction. According to this law, the emf induced in a wire is equal to the rate of change of magnetic flux through the loop of the wire.
The magnetic flux through the loop is the product of the magnetic field strength and the area of the loop. In this case, since the wire is moving horizontally to the North, the area of the loop is equal to the length of the wire multiplied by the distance it moves vertically upward. The magnetic field strength is the horizontal component of the earth's magnetic field since it is perpendicular to the direction of motion of the wire.
Let's calculate the magnetic flux first:
Magnetic flux = (Magnetic field strength) x (Area of the loop)
= (25 μT) x (50 cm x 2 m/s)
= 25 μT x 1 m x 2 m/s
= 50 μT.m²/s
Now, we can calculate the rate of change of magnetic flux:
Rate of change of magnetic flux = (Change in magnetic flux) / (Change in time)
= (50 μT.m²/s) / (1 s)
= 50 μT.m²/s
Finally, the induced emf is equal to the rate of change of magnetic flux:
Emf = Rate of change of magnetic flux
= 50 μT.m²/s
However, the given answer choices are in microvolts (μV), not in microtesla (μT). So, we need to convert the answer to microvolts:
1 μT.m²/s = 1 μV
Therefore, the emf between the ends of the wire is 50 μV, which is equal to 50 μV.
The correct answer choice is B) 50 μV.