Answer:
[tex] \vec{F}=0.40176 N \hat{k} [/tex]
Explanation:
To calculate the force we need to use this equation
[tex]\vec{F}=\int\limits^L_0 {i(\vec{dl}\times\vec{B})}[/tex]
where L is the total length of the wire
So in this case the small element of current is
[tex] \vec{dl} = dx \hat{i} [/tex]
Because x is the direction of the current flow.
As is said in the problem B is such that
[tex] \vec{B} = B \hat{j} = 0.62\hat{j} [ T][/tex]
so to use the equation above we first calculate the following cross product:
[tex]\vec{dl}\times\vec{B}=dx \hat{i}\times B \hat{j} = Bdx\hat{k}[/tex]
so the force:
[tex] F = \int\limits^L_0 {i(\vec{dl}\times\vec{B})}=\int\limits^L_0{iBdx\hat{k}} [/tex]
So here we use the fact that B=0 in any point of the x axis that is not [tex] x^{'}=0.27 [m] [/tex], that means that we only need to do the integration between a very short distant behind the point [tex] x^{'}=0.27 [m] [/tex] and a very short distant after that point, meaning:
[tex]\vec{F}= \lim_{h \to 0}{\int\limits^{x^{'}+h}_{x^{'}-h}{iBdx\hat{k}} }[/tex]
so is the same as evaluating [tex]iBx[/tex] at [tex]x=x^{'}[/tex]
that is:
[tex] 2,4 A * 0,62 T * 0,27 m \hat{k}[/tex]
[tex] 2,4 A * 0,62 (\frac{Kg}{A s^{2}}) * 0,27 m \hat{k}[/tex]
[tex] 2,4*0,62*2,7 ( \frac{ kgm }{ s^{2} } ) \hat{k}[/tex]
[tex] \vec{F}=0.40176 N \hat{k} [/tex]
