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Moving rod and induced EMF

  • Thread starter DottZakapa
  • Start date
65
4
Homework Statement
A 25.0 cm long metal rod lies in the xy-plane and makes an angle of 36.9 with the positive x-axis and an angle of 53.1 with the positive y-axis. the rod is moving in the +x-direction with a speed of 6.80 m/s. the rod is in a uniform magnetic field B= (0.120T)i-(0.220T)j-(0.0900T)k .
What is the magnitude of the emf induced in the rod? indicate in sketch which end of the rod is at higher potential.
Homework Equations
induced emf
Homework Statement: A 25.0 cm long metal rod lies in the xy-plane and makes an angle of 36.9 with the positive x-axis and an angle of 53.1 with the positive y-axis. the rod is moving in the +x-direction with a speed of 6.80 m/s. the rod is in a uniform magnetic field B= (0.120T)i-(0.220T)j-(0.0900T)k .
What is the magnitude of the emf induced in the rod? indicate in sketch which end of the rod is at higher potential.
Homework Equations: induced emf

emf =VxBxL
i have done first
vxB= (6.8 m/s i) x( (0.120T)i-(0.220T)j-(0.0900T)k)= (0.612j -1.496k)
next i split in components the length
0,25(cos 36.9 + sin 36.9)= 0.199 i+0.15j
then
(VxB)x(0.199 i+0.15j)=(0.612j -1.496k)x(0.199 i+0.15j)= (0.224i-0.297j-0.121k)

now getting the magnitude of this i don't get the right result. Can anyone explain to me why?
 

Doc Al

Mentor
44,803
1,064
emf =VxBxL
Don't take the cross-product with L. Take the scalar product of ##\vec{v}\times \vec{B}## with ##\vec{L}##. (Note that EMF is a scalar.)
 
65
4
Don't take the cross-product with L. Take the scalar product of ##\vec{v}\times \vec{B}## with ##\vec{L}##. (Note that EMF is a scalar.)
Screen Shot 2019-08-28 at 19.18.21.png


this is how it should be solved, but i don't get why is so :oldconfused:
 

Doc Al

Mentor
44,803
1,064
Is there a step you're not getting? The first step (labeled "Identify" in the given solution) is key and it's just what I had stated earlier. That's what you were doing wrong.

What textbook are you using?
 
65
4
What textbook are you using?
is a previous years exam solution
Is there a step you're not getting? .
what i am not getting is from this point on, the second passage doesn't match the first
Screen Shot 2019-08-28 at 20.31.14.png
 

Doc Al

Mentor
44,803
1,064
Sorry, but I'm not seeing which passages don't match.
 
65
4
Sorry, but I'm not seeing which passages don't match.
it starts with this
Screen Shot 2019-08-28 at 22.09.12.png

then it splits ##\vec L## which is the length of the rod in vector components as follows
Screen Shot 2019-08-28 at 22.09.31.png

you said to not take the scalar product with ##\vec L##, so why is the length in vector components?how do you get this result below? this is what i am not getting. Where is (-1.496 ##\frac V m ##) ## \hat k ## gone?
Screen Shot 2019-08-28 at 22.09.42.png


what is that sin(36.9) doing there?
 

Doc Al

Mentor
44,803
1,064
you said to not take the scalar product with ##\vec L##, so why is the length in vector components?
No, I said not to take the cross product with ##\vec L##. (Which is what you were doing in your first post.) But do take the scalar product!
how do you get this result below? this is what i am not getting. Where is (-1.496 ##\frac V m ##) ## \hat k ## gone?
Since ##\vec L## has no ## \hat k ## component, that term vanishes after taking the scalar product.
what is that sin(36.9) doing there?
That's from taking the ## \hat j ## component of ##\vec L##. Note that only the ## \hat j ## components survive the scalar product.
 
65
4
No, I said not to take the cross product with ##\vec L##. (Which is what you were doing in your first post.) But do take the scalar product!

Since ##\vec L## has no ## \hat k ## component, that term vanishes after taking the scalar product.

That's from taking the ## \hat j ## component of ##\vec L##. Note that only the ## \hat j ## components survive the scalar product.
Damn now I understood where my problem is.
I had completely forgot how a scalar product works with vectors components.
I've just recalled this i⋅j=i⋅k=j⋅k=0 and i⋅i=j⋅j=k⋅k=1
😓
Now is all bright n clear.
Sorry man, thanks for your time and patient. very gentle.
 

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