Question

A wire is oriented along the x-axis. It is connected to two batteries, and a conventional current of 2.4 A runs through the wire, in the +x direction. Along 0.27 m of the length of the wire there is a magnetic field of 0.62 tesla in the +y direction, due to a large magnet nearby. At other locations in the circuit, the magnetic field due to external sources is negligible. What is the magnitude of the magnetic force on the wire?

Answer 1

Answer:

$\vec{F}=0.40176 N \hat{k}$

Explanation:

To calculate the force we need to use this equation

$\vec{F}=\int\limits^L_0 {i(\vec{dl}\times\vec{B})}$

where L is the total length of the wire

So in this case the small element of current is

$\vec{dl} = dx \hat{i}$

Because x is the direction of the current flow.

As is said in the problem B is such that

$\vec{B} = B \hat{j} = 0.62\hat{j} [ T]$

so to use the equation above we first calculate the following cross product:

$\vec{dl}\times\vec{B}=dx \hat{i}\times B \hat{j} = Bdx\hat{k}$

so the force:

$F = \int\limits^L_0 {i(\vec{dl}\times\vec{B})}=\int\limits^L_0{iBdx\hat{k}}$

So here we use the fact that B=0 in any point of the x axis that is not $x^{'}=0.27 [m]$, that means that we only need to do the integration between a very short distant behind the point $x^{'}=0.27 [m]$ and a very short distant after that point, meaning:

$\vec{F}= \lim_{h \to 0}{\int\limits^{x^{'}+h}_{x^{'}-h}{iBdx\hat{k}} }$

so is the same as evaluating $iBx$ at $x=x^{'}$

that is:

$2,4 A * 0,62 T * 0,27 m \hat{k}$

$2,4 A * 0,62 (\frac{Kg}{A s^{2}}) * 0,27 m \hat{k}$

$2,4*0,62*2,7 ( \frac{ kgm }{ s^{2} } ) \hat{k}$

$\vec{F}=0.40176 N \hat{k}$