0
$\begingroup$

Suppose I have a closed circular loop kept on a horizontal plane with current $i$ flowing through it in a clockwise direction. A magnetic field $B$ is created by bringing a bar magnet with north pole facing the loop.enter image description here If I consider a small element $dl$ on the loop, the force acting on it due to the magnetic field would be $idlB$ in the outward direction. Similarly, all the small elements on the loop will experience a force in the outward direction which will eventually cancel out, and the resultant will be $0$.
Due to the clockwise current in the loop, a magnetic field will be generated which will make the loop act as a bar magnet with north pole facing outwards. Shouldn't the north poles(one which is created due to current and other due to bar magnet creating magnetic field $B$) repel each other? Wouldn't this make the net force on the loop non-zero? (assuming gravity-free space and no normal reaction forces). So why do we say that net force on a current-carrying loop is zero?

$\endgroup$
5
  • $\begingroup$ Magnetic field lines "interact" with each other? can you give please an example? $\endgroup$ Sep 8 at 16:49
  • $\begingroup$ A magnetic field 𝐵 exists perpendicular to the plane of the loop in the inward direction how can this possibly be in the "inward" direction if it's perpindicular to the plane of the loop? $\endgroup$
    – Señor O
    Sep 8 at 16:50
  • $\begingroup$ i have editted my question making it more clear. please check it further $\endgroup$
    – nimit jain
    Sep 8 at 17:05
  • $\begingroup$ As you've shown, both the net force and net torque on the loop is zero, but imagine if the loop was at a slight angle to the magnetic field (it would help to use a square loop to simplify the analysis). $\endgroup$
    – Triatticus
    Sep 8 at 17:44
  • $\begingroup$ A bar magnet in uniform magnetic field experiences no net force. So, your expectation of a net force is not justified. $\endgroup$
    – nasu
    Sep 8 at 21:26
1
$\begingroup$

(a) What you are forgetting is that the field beyond the North Pole of a bar magnet is not parallel to the magnet's longitudinal axis of symmetry, but has a component directed outwards from this axis. The magnetic field lines splay out! [The exception is along the axis of symmetry itself.]

So the motor effect force on your loop element, $dl$, will have a component parallel to the (common) axis of the magnet and loop. So we do have agreement with the loop experiencing a force as would another magnet. The force on the loop depends on the actual magnet's field being non-uniform. [In a uniform magnetic field the loop would experience no net force but perhaps (depending on orientation) a torque.]

(b) "Due to the clockwise current in the loop, a magnetic field will be generated which will make the loop act as a bar magnet with north pole facing outwards." I don't know what you mean by 'facing outwards' The North pole of the loop is the face of the loop further from the viewer of your screen. This is because the magnetic field lines are coming out of this face.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.