Calculating EMF in Steel Beam Impact Before Landing

[xswtxoj]

Homework Statement

A 11.2m long steel been is dropped from a height of 9.51m. The horizontal component of the Earth's over the region is 18.5x10^-6T. What's the emf in the beam impact just before the impact with the earth, assuming its long dimension remains in a horizontal plane, oriented perpendicularly to the horizontal component of the Earth's magnetic field?

Homework Equations

e= mag flx/t

The Attempt at a Solution

A= 1/2*b*h
A=1/2 x 11.2 x9.51= 53.36

18.5e-6* 53.26
but there's no time

 

[Redbelly98]

Homework Statement

A 11.2m long steel been is dropped from a height of 9.51m. The horizontal component of the Earth's over the region is 18.5x10^-6T. What's the emf in the beam impact just before the impact with the earth, assuming its long dimension remains in a horizontal plane, oriented perpendicularly to the horizontal component of the Earth's magnetic field?

Homework Equations

e= mag flx/t

The Attempt at a Solution

A= 1/2*b*h
A=1/2 x 11.2 x9.51= 53.36

That would be the area of a triangle, but I fail to see the relevance?

18.5e-6* 53.26
but there's no time

My own hunch is that the electric and magnetic forces on a test charge Q in the beam would have to be equal in magnitude, opposite in direction.

So, what is the magnetic force on a charge Q?

 

[nlightner]

given in the problem

I would like to clarify a few things before providing a response. Firstly, I assume that the problem is referring to the electromagnetic force (EMF) in the steel beam due to its motion through Earth's magnetic field. Secondly, I will also assume that the steel beam is a solid, rectangular prism with a cross-sectional area of 53.36 square meters and a length of 11.2 meters.

Now, to calculate the EMF in the steel beam, we need to use the equation e = vBL, where e is the EMF, v is the velocity of the beam, B is the magnitude of the horizontal component of the Earth's magnetic field, and L is the length of the beam. Since the beam is dropped from a height of 9.51 meters, we can use the equation v = √(2gh) to find the velocity of the beam just before impact, where g is the acceleration due to gravity (9.8 m/s^2) and h is the height from which the beam is dropped.

So, the velocity of the beam just before impact is v = √(2*9.8*9.51) = √186.96 = 13.67 m/s.

Now, plugging in the values for v, B, and L, we get: e = (13.67 m/s)(18.5x10^-6 T)(11.2 m) = 2.82x10^-3 V.

Therefore, the EMF in the steel beam just before impact is 2.82 millivolts. It is important to note that this calculation assumes that the beam is moving in a straight line and that the Earth's magnetic field is uniform over the entire region. In reality, the beam would experience some deflection due to the Earth's magnetic field, and the strength of the field may vary over the region. These factors may affect the accuracy of the calculated EMF. Additionally, the time of impact is not provided, so we cannot accurately calculate the time in which the EMF is present in the beam.