## Magnetic Field of Short-Straight Conductors

- Category: Enrichments
- Written by fisikastudycenter

physics.fisikastudycenter.com - Magnetic field of a Straight Current Carrying Conductor with Certain of Length, high school physics grade 12. Commonly cases in high school physics of straight current-carrying conductors magnetic field are for very long conductors. So we use this equation to calculate the magnetic fields:

where

B = magnetic field at a point (Tesla)

i = current flows (A)

a = the distance of the point from the conductor (m)

μ_{o} = 4π ⋅ 10^{−7} Wb/A m

In above case the we assume that the conductor is very long. In another cases we are given a certain lenght of conductors or short conductors, so we don't use the previous equation anymore. Here is the equation that we should use:

with α and β are the angles formed between point C and the conductor, show the figure above.

Follow the problem sample below to get more understanding:

A current i of 2 A flows along conductor PU as below figure. The length of QR, RS and ST are the same that are 10 cm. Determine the magnitude of magnetic field produced at point C!

For simplification, assume there are three conductors producing magnetic field at point C. Conductor QR, conductor RS and conductor ST. We should ignore conductor PQ and TU that do not produce magnetic field at point C. The diagram of conductor QR and ST is shown below:

Use phytagorean theorem to find 5√5 cm as mentioned in the figure. Then state the values of sin α and sin β for each conductor. Starting for QR conductor:

Repeat the steps for ST conductor and we'll find the same value of B_{ST} = B_{QR}

Below the calculation for conductor RS

You shouldn't get difficulties how to find the 5√5 cm and the values of sin α and sin β of conductor SR

Finally use the right hand rules to find the directions of the three magnetic fields at point C, then we get the three magnetic fields have the same in directions, into paper field, so just adding the three: