The Magnitude of Induced EMF in a Moving Rod?

[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.

Relevant 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]

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]

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]

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]

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]

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.