A 0.80-m aluminium bar is held with its length parallel to the east-west direction and dropped from a bridge. Just before the bar hits the river below, its speed is 22 m/s, and the emf induced across its length is 5 X 10-4 V. Assuming the horizontal component of the earth's magnetic field at the location of the bar points directly north, (a) determine the magnitude of the horizontal component of the earth's magnetic field, and (b) state whether the east end or the west end of the bar is positive.

Question

A 0.80-m aluminium bar is held with its length parallel to the east-west direction and dropped from a bridge. Just before the bar hits the river below, its speed is 22 m/s, and the emf induced across its length is 5 X 10-4 V. Assuming the horizontal component of the earth's magnetic field at the location of the bar points directly north, (a) determine the magnitude of the horizontal component of the earth's magnetic field, and (b) state whether the east end or the west end of the bar is positive.

Answer 1

5.9*10|^42

Answer 2

To solve this problem, we will use Faraday's law of electromagnetic induction, which states that the induced electromotive force (emf) is directly proportional to the rate of change of magnetic flux through a surface. The equation for induced emf can be written as:

emf = -N * (ΔΦ/Δt)

Where:
emf is the induced electromotive force
N is the number of turns in the coil (which we can assume to be 1 in this case, as there is a single bar being dropped)
ΔΦ is the change in magnetic flux
Δt is the time interval over which the change occurs

In this problem, the bar is falling vertically downwards, so its length is the only dimension that will change. The magnetic flux is given by the product of the magnetic field strength (B) and the area (A) perpendicular to the magnetic field. In this case, the bar is falling parallel to the magnetic field, so the area is equal to the length of the bar.

Now let's find the magnitude of the horizontal component of the Earth's magnetic field (B).

emf = -N * (B * A) / Δt

In this case, the induced emf is given as 5 x 10^-4 V, the length of the bar is 0.80 m, and the speed of the bar can be considered as the time it takes to change the length (Δt) as it falls. So, we have:

5 x 10^-4 V = -1 * (B * 0.80 m) / 22 m/s

Now, let's solve for B:

B = (5 x 10^-4 V) * (22 m/s) / (-0.80 m)

B ≈ -0.0138 T

The negative sign indicates that the horizontal component of the Earth's magnetic field is pointing in the opposite direction to the assumed north direction.

Now, let's determine which end of the bar is positive. According to the right-hand rule, if the magnetic field is pointing north and the bar is falling downwards, the induced current in the bar will create a magnetic field opposing the change in flux. This means that the west end (left end) of the bar will be positive.