Calculating Induced EMF in a 5.56m Steel Beam Dropped from 3.97m

Click For Summary
To calculate the induced EMF in a 5.56m steel beam dropped from a height of 3.97m, the key factors include the beam's orientation relative to the Earth's magnetic field and the acceleration due to gravity. The horizontal component of the Earth's magnetic field is 28.4e-6T, and gravity accelerates the beam at 9.8 m/s². The induced EMF can be determined using the equation E=Blv, where velocity (v) can be found using kinematic equations. Specifically, the final velocity just before impact can be calculated with v_f² = v_i² + 2a(y_f - y_i). Understanding these principles allows for the accurate calculation of the induced EMF just before the beam strikes the ground.
Aerospace
Messages
39
Reaction score
0
A 5.56m long steel beam is accidentally dropped by a construction crane from a height of 3.97m. The horizontal component of the Earth's magnetic field over the region is 28.4e-6T. Acceleration of gravity is 9.8 m/s^2.
What is the induced emf in the beam just before 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?

I sort of have an idea how to begin but the whole gravity thing is confusing me. Any help?
 
Physics news on Phys.org
I think that gravity only plays a role since you will want to know how fast the beam is traveling through the magnetic field just before it hits the ground.
 
YUP! That's what I was thinking too. One of the equations for induced emf is E=Blv but how do I get the velocity from the gravity because i am not given a time period. It's probably something just basic...but I can't seem to grasp on it at this moment.
 
Just use the kinematic equations. You are given the height it is falling from, and you know how fast it accelerates due to gravity.
Without time, use v_{f}^2=v_{i}^2+2a(y_{f}-y_{i})
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 41 ·
2
Replies
41
Views
8K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 9 ·
Replies
9
Views
5K
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K