Moving rod and induced EMF

DottZakapa

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?

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Doc Al

Mentor
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.)

DottZakapa

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.) this is how it should be solved, but i don't get why is so Doc Al

Mentor
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?

DottZakapa

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 Doc Al

Mentor
Sorry, but I'm not seeing which passages don't match.

DottZakapa

Sorry, but I'm not seeing which passages don't match.
it starts with this then it splits $\vec L$ which is the length of the rod in vector components as follows 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? what is that sin(36.9) doing there?

Doc Al

Mentor
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.

• DottZakapa

DottZakapa

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.

"Moving rod and induced EMF"

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