Force and torque on a wire carrying a current

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theKeeblerElf
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Homework Statement



A wire along the x-axis carries current I in the negative x-direction through the magnetic field
[tex]\vec{}B[/tex] = B0*(x/L) [tex]\hat{}k[/tex] 0[tex]\leq[/tex]x[tex]\leq[/tex]L
= 0 elsewhere

Part a was to draw a graph of B versus x over the interval -L[tex]\leq[/tex]x[tex]\leq[/tex]L, which I did.

b. Find an expression for the net force F_{\rm net} on the wire. Express your answer in terms of the variables I, L, and B0

c. Find an expression for the net torque on the wire about the point x = 0.
Express your answer in terms of the variables I, L, and B0

Homework Equations



Fnet=I*L*B*sin[tex]\alpha[/tex]
[tex]\tau[/tex]net=I*L2*B*sin[tex]\alpha[/tex]

The Attempt at a Solution



I thought that because I is carried in the -[tex]\hat{}i[/tex] direction and B points in the [tex]\hat{}k[/tex] direction that [tex]\alpha[/tex]=90 degrees, meaning I would be multiplying by 1. However, for both parts, when I submit I*L*B0 and I*L2*B0, respectively, the website says "Your answer either contains an incorrect numerical multiplier or is missing one."

For part b I've tried submitting -I*L*B0 because actually calculating I[tex]\hat{}L[/tex] X [tex]\hat{}B[/tex] yields a negative answer, but that didn't work either.

I've also tried to use x in my answer, and the website says the correct answer doesn't depend on it.

Any help would be much appreciated!
 
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theKeeblerElf said:
Fnet=I*L*B*sin[tex]\alpha[/tex]
[tex]\tau[/tex]net=I*L2*B*sin[tex]\alpha[/tex]

This is only true when the magnetic field is constant over the length of the wire. More generally, you will have to divide the wire into pieces [itex]d\textbf{l}[/itex] so small that the magnetic field is effectively constant/uniform over the length of the piece, calculate the force on each piece:

[tex]d\textbf{F}=Id\textbf{l}\times\textbf{B}[/tex]

and then add up (integrate) all these little forces:

[tex]\textbf{F}=\int Id\textbf{l}\times\textbf{B}[/tex]

You must do something similar for the torque.