Magnetic force

**johett** · Mar 24th 2008, 06:23 AM

Q: A horizontal 6.0m long wire that runs from west to east is in a 0.03-T magnetic field with a direction that is northeast

- If a 4.5A current flows east through the conductor, what is the magnitude and direction of the force on the wire?

Q: Copper metal wire has a linear density of 0.010kg/m. A sample of this wire is stretched horizontally in an area where the horizontal component of Earth's magnetic field of strength 2.0x10^-5T passes through the wire at right angles.

- What current must be applied to the wire if the weight of the entire wire is supported by the magnetic force?

**iknowone** · Mar 24th 2008, 11:44 AM

There is a simple equation for solving for the force on a current carrying wire in a magnetic field:

$F = aL \times B$

where $a$ is the current in the wire (amps), L is the vector representation of the direction of the current with length that of the wire, and B is the magnetic field vector.

$F = 4.5 \left( \begin{array}{c} 6 \\ 0 \\ 0 \end{array} \right) \times \left( \begin{array}{c} .02121 \\ .02121 \\ 0 \end{array} \right)$

For the second problem we have a force diagram on the wire with 2 vectors that we want to make equal. The force downward due to gravity acting on the wire and the Lorentz force force upward. It points upward since the horizontal wire is oriented perpendicular to the horizontal component of the magnetic field and the cross product of two vectors is a third vector perpendicular to both vectors.

The force due to gravity is $9.81m$ where $m$ is the mass of the wire in kilograms. The mass is given by $.01x$ where $x$ is the length of the wire in meters.

The equation becomes:

$ (.01x)\left( \begin{array}{c} 0 \\ 0 \\ -9.81 \end{array} \right) = a\left( \begin{array}{c} 0 \\ x \\ 0 \end{array} \right) \times \left( \begin{array}{c} 2.0\times 10^{-5} \\ 0 \\ 0 \end{array} \right)$

Solving for $a$ .

$-.0981 x = -.00002 ax \Rightarrow a = 4900.5\text{A}$ .

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