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Force Between Two Parallel Current Carrying Conductor


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Published on:  on 10th Apr, 2018

Description:

 
Please note, this is a STATIC archive of website www.tutorialspoint.com from 11 May 2019, cach3.com does not collect or store any user information, there is no "phishing" involved.

Two current carrying straight conductors placed near each other will exert (magnetic) forces on each other due to magnetic field of each other.

Parallel Current

Figure shows two long parallel conductors separated by a distance d and bearing currents ia and ib. The conductor ‘a’ produces a magnetic field Ba at all points along the conductor ‘b’. The right-hand rule tells us that the direction of this field is downwards. Its magnitude is given by

Ba =
μ0ia/2πd

The conductor ‘b’ will experience a sideways force on account of the external field Ba. The direction of this force is towards the conductor ‘a’. you can verify this either by the cross product rule of vectors or by Fleming’s left hand rule. We label this force Fba, the force on a segment L of ‘b’ due to ‘a’. The magnitude of this force is given by,

Fba = iaIBa

Fba/L
=
μ0iaib/2πd
 

From considerations similar to above we can find that the force Fab, on a segment of length L of ‘a’ due to the current in ‘b’ is equal in magnitude to ‘b’ and directed towards ‘b’.

Note −  Parallel current carrying wires attract, and anti-parallel current carrying wires repel each other.


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