Consider a circular current-carrying coil.
What is the ratio of the magnetic field at the centre of a circular coil of radius R carrying a current to that at a point on its axis at a distance R/2 from the centre?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 5\sqrt{5}:8 \)
For a circular loop of radius \( R \) carrying current \( I \), the field on its axis at a distance \( x \) from the centre is
\( \displaystyle B_x=\frac{\mu_0IR^{2}}{2\,(x^{2}+R^{2})^{3/2}} \)
At the centre (\( x=0 \)) this becomes \( B_0=\mu_0I/2R \).
At \( x=R/2 \), the bracket is:
\( \begin{aligned} x^{2}+R^{2} &= \frac{R^{2}}{4}+R^{2} \\ &= \frac{5R^{2}}{4} \end{aligned} \)
\( \begin{aligned} (x^{2}+R^{2})^{3/2} &= \left(\frac{5R^{2}}{4}\right)^{3/2} \\ &= \frac{5\sqrt{5}\,R^{3}}{8} \end{aligned} \)
The field on the axis is then:
\( \begin{aligned} B_x &= \frac{\mu_0IR^{2}}{2}\times\frac{8}{5\sqrt{5}\,R^{3}} \\ &= \frac{8}{5\sqrt{5}}\cdot\frac{\mu_0I}{2R} \end{aligned} \)
The ratio of the two fields is:
\( \begin{aligned} \frac{B_0}{B_x} &= \frac{\mu_0I/2R}{\left(8/5\sqrt{5}\right)\,\mu_0I/2R} \\ &= \frac{5\sqrt{5}}{8} \end{aligned} \)
Therefore, the ratio is \( 5\sqrt{5}:8 \).