I know that many questions have been asked regarding this, but browsing over the questions, none of them seem to clarify my doubts.

By applying Ampere's circuital law to a current-carrying circular coil, choosing a circle of radius $a$ which is smaller than the radius of the coil, and integrating over it, I found that magnetic field inside a current carrying coil is zero since there is no current enclosed within the circle. So the rhs in the equation $\oint B\cdot dl = \mu_0 I$ becomes zero which effectively means the magnetic field is zero. However, we already know that by using Biot-Savart's law we can easily derive that magnetic field at the center of a current-carrying circular coil is $B = \frac{\mu_0 I}{2r}$.

How are these two contradictory? If not, have I applied the two laws correctly? How can we derive the magnetic field inside a current carrying circular coil using Ampere's circuital law?

  • 2
    $\begingroup$ Just because an integral involving $B$ is zero does not imply $B$ is zero. $\endgroup$ – knzhou Jul 30 '16 at 0:30
  • $\begingroup$ then it should imply that the dot product is zero.but it doesnot matter whatever the curve be the integral is always zero.it therefore should imply that B is zero. $\endgroup$ – Pink Jul 30 '16 at 0:33
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    $\begingroup$ This is wrong. For example, think back to classical mechanics. The integral of any conservative vector field is zero. That doesn't mean all conservative vector fields are zero. $\endgroup$ – knzhou Jul 30 '16 at 0:35
  • $\begingroup$ then what about a long torroidal solenoid.in this case only,we should not come to the conclusion that B is zero inside the torroid not enclosing any current. $\endgroup$ – Pink Jul 30 '16 at 0:58
  • $\begingroup$ it isn't. it's just pretty small. $\endgroup$ – knzhou Jul 30 '16 at 0:59

The problem lies in the fact that you are computing the path integral that forms the LHS of Ampeere's law incorrectly. We know that for a long solenoid, the magnetic field lines are mostly confined to the center of the coil and PARALLEL to its axis. If I understand correctly, your circle is perpendicular to this axis. Granted, you could then rotate the circle 90 degrees, but evaluating this would still be tedious, since the field lines are straight lines and your path is still a circle.

To take full advantage of the geometry of the situation, the ideal method is to consider the integral path to be a square- with one side parallel to the solenoid axis and within the coil, the opposite side outside the coil, and , as follows from the shape of a square, the other two sides perpendicular to the solenoid axis and the magnetic field. This square will ,of course, contain some current carrying segments of the coil, so the RHS of Ampere's law is no longer zero; as it should be.

  • $\begingroup$ if the rhs itself becomes zero why should i go for lhs $\endgroup$ – Pink Jul 30 '16 at 0:19
  • $\begingroup$ the rhs doesn't become zero, thats what I'm trying to tell you $\endgroup$ – Luke B-hyndU Jul 30 '16 at 0:21
  • $\begingroup$ but inside a current carrying circular coil,there is no current enclosed by any any closed curve we may imagine so that rhs will always reduce to zero.this means that everywhere inside a current carrying circular coil B is zero.which is contradictory. $\endgroup$ – Pink Jul 30 '16 at 0:24
  • $\begingroup$ consider a closed curve that isn't completely contained within the coil $\endgroup$ – Luke B-hyndU Jul 30 '16 at 0:34
  • $\begingroup$ why should i do that and how can it be helpful to me in calculating the magnetic field at the centre of the current carrying circular coil using ampere's law. $\endgroup$ – Pink Jul 30 '16 at 0:38

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