91Ó°ÊÓ

According to Biot-Savart law, the magnetic field due to a current I in a small element dl at a point P at distance r is given by \(dB= \frac { \mu I dl sin \theta }{ 4 \pi r^2 }\), where\( \mu \)= Permeability of free spaceI = Currentdl = Small section of wirer = Distance of point P from dl\( \theta \) = Angle between dl and line joining dl and point PIn this case, the direction of dl and r is same, so \( \theta \) = 0Therefore the magnetic field due to a small element dl at the axis of the ring at distance z from the centre of the ring will be \( dB= \frac { \mu I dl }{ 4 \pi (a^{2} + z^{2}) }\).The total magnetic field at the point P due to the entire ring can be obtained by integrating from 0 to 2\(\pi\) as \( B= \frac { \mu I a^2 }{ 2 (a^{2} + z^{2})^{3/2} } \). This will give us the magnetic field due to one ring. Since both rings are producing magnetic fields in opposite directions, we have to subtract them.

The magnetic field at a point on the axis at distance z from the center of the upper ring due to the upper ring is \( B_{1}= \frac { \mu I a^2 }{ 2 (a^{2} + (z - \frac{ \epsilon }{2})^{2})^{3/2} }\) and due to the lower ring is \( B_{2}= \frac { \mu I a^2 }{ 2 (a^{2} + (z + \frac{ \epsilon }{2})^{2})^{3/2} }\). The total field \( B = B_{1} - B_{2} \).

We need to equate the derivative of the total magnetic field \( B \) with respect to \( z \) to zero. This ensures that we're finding the maximum or minimum of the function. Therefore, \( \frac{ dB }{ dz } = 0 \).Upon calculating, we find \( z = \frac{ \epsilon }{2} \).

After finding the maximum magnetic field point, it's important to verify that this point corresponds to a maximum. This can be done by checking the value of the second derivative \( \frac{ d^{2}B }{ dz^{2} } \) at \( z = \frac{ \epsilon }{2} \). If the second derivative is negative, then we are at a maximum point. Indeed, the calculation confirms this.