If \(B_1\) is the magnetic field induction at a point on the axis of a circular coil of radius \(R\)…
If \(B_1\) is the magnetic field induction at a point on the axis of a circular coil of radius \(R\) situated at a distance \(R \sqrt{3}\) and \(B_2\) is the magnetic field at the centre of the coil, then the ratio of \(\frac{B_1}{B_2}\) is equal to
\(\frac{1}{3}\)
\(\frac{1}{8}\)
\(\frac{1}{4}\)
\(\frac{1}{2}\)
Solution
According to question,
\(B_1=\frac{\mu_0 I R^2}{2\left(R^2+r^2\right)^{3 / 2}}\)
Here, \(\quad r=R \sqrt{3}\)
\(\begin{aligned}
\therefore \quad B_1 & =\frac{\mu_0 I R^2}{2\left[R^2+(R \sqrt{3})^2\right]^{3 / 2}}=\frac{\mu_0 I R^2}{2\left[4 R^2\right]^{3 / 2}}=\frac{\mu_0 I R^2}{16 R^3} \\
B_1 & =\frac{\mu_0 I}{16 R} \quad \ldots (i)
\end{aligned}\)
and \(\quad B_2=\frac{\mu_0 I}{2 R}\) ...(ii)
[magnetic field at centre]
Now, dividing Eq. (i) by Eq. (ii), we get
\(\therefore \quad \frac{B_1}{B_2}=\frac{\mu_0 I / 16 R}{\mu_0 I / 2 R}=\frac{1}{8}\)