Question

For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its centre is given by,
\[B=\dfrac{{{\mu }_{0}}I{{R}^{2}}N}{2{{\left( {{x}^{2}}+{{R}^{2}} \right)}^{{}^{3}/{}_{2}}}}\]​

(a) Show that this reduces to the familiar result for the field at the centre of the coil.

(b) Consider two parallel coaxial circular coils of equal radius R, and number of turns N, carrying equal currents in the same direction, and separated by a distance R. Show that the field on the axis around the midpoint between the coils is uniform over a distance that is small as compared to R, and is given by. \[B=0.72\dfrac{{{\mu }_{0}}NI}{R}\]​
[Such an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.]

Answer 1

Hint: At the center, the value of x will be zero. For the second part, calculate the magnetic field at a point near to the midpoint due to both coils. Now, according to the direction of current flow in each coil, write the expression for the resultant magnetic field at that point.

Complete step by step solution:
Given,
Radius of coil = R
Number of turns = N
Current flowing through the coil = I
Magnetic field at a point on the axis at a distance x is given as

\[B=\dfrac{{{\mu }_{0}}I{{R}^{2}}N}{2{{\left( {{x}^{2}}+{{R}^{2}} \right)}^{{}^{3}/{}_{2}}}}\]

(a) At the center of the coil, x=0, therefore, the magnetic field at the center is given as

\[\begin{align}
  & {{B}_{C}}=\dfrac{{{\mu }_{0}}I{{R}^{2}}N}{2{{\left( 0+{{R}^{2}} \right)}^{{}^{3}/{}_{2}}}} \\
 & {{B}_{C}}=\dfrac{{{\mu }_{0}}I{{R}^{2}}N}{2{{R}^{3}}} \\
 & {{B}_{C}}=\dfrac{{{\mu }_{0}}IN}{2R} \\
\end{align}\]
Thus, the given expression reduces to the familiar result for the field at the centre of the coil.

(b) Consider two parallel coaxial circular coils X and Y of equal radius R, and the number of turns N, carrying equal currents I in the same direction, and separated by a distance R as shown in the figure. Let O be the midpoint on the axis between two coils. Let us assume we have to find the magnetic field on the axis at a distance d (d << R) from the center. We have to show this field is uniform and has the value \[B=0.72\dfrac{{{\mu }_{0}}NI}{R}\]

Let ${{B}_{X}}$ the magnetic field at a point Q due to coil X. It can be calculated as
\[{{B}_{X}}=\dfrac{{{\mu }_{0}}I{{R}^{2}}N}{2{{\left( {{(\dfrac{R}{2}+d)}^{2}}+{{R}^{2}} \right)}^{{}^{3}/{}_{2}}}}\]
Similarly, let ${{B}_{Y}}$ the magnetic field at a point Q due to coil Y. It can be calculated as
\[{{B}_{Y}}=\dfrac{{{\mu }_{0}}I{{R}^{2}}N}{2{{\left( {{(\dfrac{R}{2}-d)}^{2}}+{{R}^{2}} \right)}^{{}^{3}/{}_{2}}}}\]
As the current in both magnetic fields flows in the same direction, the net magnetic field at point Q is given as

$B={{B}_{X}}+{{B}_{Y}}$

Substituting the values of ${{B}_{X}}$ and ${{B}_{Y}}$ in the above equation, we get,

$B=\dfrac{{{\mu }_{0}}I{{R}^{2}}N}{2{{\left( {{(\dfrac{R}{2}+d)}^{2}}+{{R}^{2}} \right)}^{{}^{3}/{}_{2}}}}+\dfrac{{{\mu }_{0}}I{{R}^{2}}N}{2{{\left( {{(\dfrac{R}{2}-d)}^{2}}+{{R}^{2}} \right)}^{{}^{3}/{}_{2}}}}$
$B=\dfrac{{{\mu }_{0}}I{{R}^{2}}N}{2}\left[ \dfrac{1}{{{\left( {{(\dfrac{R}{2}+d)}^{2}}+{{R}^{2}} \right)}^{{}^{3}/{}_{2}}}}+\dfrac{1}{{{\left( {{(\dfrac{R}{2}-d)}^{2}}+{{R}^{2}} \right)}^{{}^{3}/{}_{2}}}} \right]$
$B=\dfrac{{{\mu }_{0}}I{{R}^{2}}N}{2}\left[ \dfrac{1}{{{\left( \dfrac{{{R}^{2}}}{4}+{{d}^{2}}+Rd+{{R}^{2}} \right)}^{{}^{3}/{}_{2}}}}+\dfrac{1}{{{\left( \dfrac{{{R}^{2}}}{4}+{{d}^{2}}-Rd+{{R}^{2}} \right)}^{{}^{3}/{}_{2}}}} \right]$
$B=\dfrac{{{\mu }_{0}}I{{R}^{2}}N}{2}\left[ \dfrac{1}{{{\left( \dfrac{5{{R}^{2}}}{4}+{{d}^{2}}+Rd \right)}^{{}^{3}/{}_{2}}}}+\dfrac{1}{{{\left( \dfrac{5{{R}^{2}}}{4}+{{d}^{2}}-Rd \right)}^{{}^{3}/{}_{2}}}} \right]$
Since d << R, we neglect ${{d}^{2}}$ in the above equation.

Therefore, we will have,
$B=\dfrac{{{\mu }_{0}}I{{R}^{2}}N}{2}\left[ \dfrac{1}{{{\left( \dfrac{5{{R}^{2}}}{4}+Rd \right)}^{{}^{3}/{}_{2}}}}+\dfrac{1}{{{\left( \dfrac{5{{R}^{2}}}{4}-Rd \right)}^{{}^{3}/{}_{2}}}} \right]$

Taking $\dfrac{5{{R}^{2}}}{4}$ outside the brackets,

$B=\dfrac{{{\mu }_{0}}I{{R}^{2}}N}{2{{\left( \dfrac{5{{R}^{2}}}{4} \right)}^{{}^{3}/{}_{2}}}}\left[ \dfrac{1}{{{\left( 1+\dfrac{4d}{5R} \right)}^{{}^{3}/{}_{2}}}}+\dfrac{1}{{{\left( 1-\dfrac{4d}{5R} \right)}^{{}^{3}/{}_{2}}}} \right]$

Using binomial theorem and neglecting higher order terms, we get,

\[B=\dfrac{{{\mu }_{0}}I{{R}^{2}}N}{2{{\left( \dfrac{5{{R}^{2}}}{4} \right)}^{{}^{3}/{}_{2}}}}\left[ \left( 1-\dfrac{3}{2}\times \dfrac{4d}{5R} \right)+\left( 1+\dfrac{3}{2}\times \dfrac{4d}{5R} \right) \right]\]
\[B=\dfrac{{{\mu }_{0}}I{{R}^{2}}N}{2{{\left( \dfrac{5{{R}^{2}}}{4} \right)}^{{}^{3}/{}_{2}}}}\times 2\]
\[B={{\left( \dfrac{4}{5} \right)}^{{}^{3}/{}_{2}}}\dfrac{{{\mu }_{0}}IN}{R}\]
\[B=0.72\dfrac{{{\mu }_{0}}IN}{R}\]
We can obtain a similar expression for point P also, thus in general the field on the axis around the midpoint between the coils is uniform over a distance that is small as compared to R, and is given by. \[B=0.72\dfrac{{{\mu }_{0}}NI}{R}\]​

Note: Note that the second result is true only for small d values. If the direction of current in two coils is the same then the resultant magnetic field at a given point is just the sum of magnetic fields due to each coil. If the direction of current is opposite in two coils then the resultant magnetic field at a given point is just the difference of magnetic fields due to each coil.