Question

Wire 1, with mass 0.010kg and length 1.0m, has a square cross section and is initially at rest on a table. it is connected by flexible leads to a battery and carries a steady current of 1.5A. When wire 2 is placed parallel to wire 1 a distance of 2.0mm away, wire 1 begins to slide across the table away from wire 2. If the coefficient of static friction between wire 1 and the table is 0.050, what must be the minimum current through wire 2?

This seems kind of simple.. but I can't seem to solve it for the life of me.

Answer 1

@IrishBoy123 @Michele_Laino

Answer 2

the force moving the wire will be the force that you get between 2 parallel wires: force on moving wire (#2) will be \(\Large \frac{F_2}{l} = \frac{\mu_0I_1I_2}{2\pi d}\)

for the moving wire \(\large F= \mu R\)

Answer 3

What is \(R\) in this case?

Answer 4

I think it is the weight of wire #1, namely:
\(R=m_1 \cdot g=0.010 \cdot 10=0.1 \,newtons\)\)

Answer 5

in agreement with @IrishBoy123 the force acting on the wire #1 is the same:
\[\Large {F_1} = {F_2} = \frac{{2{I_1}{I_2}L}}{{d{c^2}}},\quad \left( {CGS} \right)\]
so the condition for motion is:
\[\Large \frac{{2{I_1}{I_2}L}}{{d{c^2}}} \geqslant \mu {m_1}g\]

Answer 6

where, as usual, \(\large c\) is the light speed in vacuum

Answer 7

@Michele_Laino That's interesting.. I'm not familiar with CGS and using the speed of light for magnetism. Could you explain a little bit how it works, or have a good link? Wikipedia seemed to confuse me a little bit more :(

Answer 8

magnetic static field, obeys to these equations:
\[\huge \begin{gathered}
{\text{div}}{\mathbf{B}} = 0 \hfill \\
{\text{rot}}{\mathbf{B}} = \frac{{4\pi }}{c}{\mathbf{J}} \hfill \\
\end{gathered} \]
where \(\Large {\mathbf{J}}\) is the vector of density current. Those equations are written using the \((CGS)\) system

Answer 9

Please note that, those equations are differential equations, namely they have been written, in differential form, which means that they have been written using partial derivatives

Answer 10

Please note that such equations have been written in differential form, namely have been written using partial derivatives. Of course we can rewrite them in integral form (with no partial derivatives)

Answer 11

for example, I consider the second equation, and I apply the theorem of \(Stokes\), so I can write this:
\[\huge {\text{circ}}{\mathbf{B}} = \frac{{4\pi }}{c}I\]
where \(\Large {\text{circ}}{\mathbf{B}} \), is the circulation of vector \(\Large {\mathbf{B}}\)along a closed path whithin there is some currents:

Answer 12

namely such path encloses some currents \(I_1,\,I_2\)

Answer 13

Example of application:
I consider an infinite rectilinear wire, and \(I\) is the current inside that wire. As we know the static magnetic field, is like below: