Force on a current carrying wire of arbitrary shape in a magnetic fieldTo obtain the net force by an external magnetic field on a current carrying wire, we divide the wire into small sections of infinitesimal lengths. The force due to the magnetic field on a given section is then given by d F = i ( d l × B ), where B is the external magnetic field at that section.Consider one of these sections. If the wire is not a straight line(has some random shape), then won't the current in a different section of the wire exert a force on the considered section due to the magnetic field produced due to that section?I am aware that when we consider any closed system, we are not concerned with the internal forces operating within the system, as internal forces, always occurring in pairs, impart no effect to the system as a whole(to its centre of mass, to be specific).However, this reasoning is only valid as long as the internal forces obey Newton's third law. But, as I have learnt, magnetic force due to charged particles DOES NOT obey Newton's third law. So, if I were to consider the current carrying wire as my system, the external magnetic field will indeed exert a force on the wire. But what about the internal forces and why are they not considered?To summarize: WHY DOES A CURRENT CARRYING WIRE OF RANDOM SHAPE NOT EXERT A FORCE ON ITSELF?Please correct me where I am wrong.

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Force on a current carrying wire of arbitrary shape in a magnetic fieldTo obtain the net force by an external magnetic field on a current carrying wire, we divide the wire into small sections of infinitesimal lengths. The force due to the magnetic field on a given section is then given by   d  F  =  i  (  d  l  ×  B  ), where B is the external magnetic field at that section.Consider one of these sections. If the wire is not a straight line(has some random shape), then won&#039;t the current in a different section of the wire exert a force on the considered section due to the magnetic field produced due to that section?I am aware that when we consider any closed system, we are not concerned with the internal forces operating within the system, as internal forces, always occurring in pairs, impart no effect to the system as a whole(to its centre of mass, to be specific).However, this reasoning is only valid as long as the internal forces obey Newton&#039;s third law. But, as I have learnt, magnetic force due to charged particles DOES NOT obey Newton&#039;s third law. So, if I were to consider the current carrying wire as my system, the external magnetic field will indeed exert a force on the wire. But what about the internal forces and why are they not considered?To summarize: WHY DOES A CURRENT CARRYING WIRE OF RANDOM SHAPE NOT EXERT A FORCE ON ITSELF?Please correct me where I am wrong.

Yasminru · Accepted Answer

Contrary to your final assertion, the wire of random shape DOES exert a force on itself. Like the external magnetic field, this force (usually quite weak) causes an outward pressure which tries to maximise the area of the loop. These forces do occur in action-reaction pairs in accordance with Newton&#039;s 3rd Law.As you acknowledge, when it comes to calculating the effect of an external magnetic field on the current loop, internal forces are ignored. It is assumed that any internal magnetic forces are already balanced by internal stresses (inter-molecular forces) in the material.The external magnetic field is a portion of a (possibly much) larger system. To apply Newton&#039;s 3rd Law to the current loop and external magnetic field we would have to identify the larger system that is generating the magnetic field. Usually the current loop is assumed to have an insignificant effect on that larger system, just as a ball thrown into the air has an insignificant effect on the Earth.So if I have understood you correctly, there is in fact no inconsistency in your question which requires an explanation.

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