Search any question & find its solution
Question:
Answered & Verified by Expert
A circular current carrying coil has a radius $\mathrm{R}$. The distance from the centre of the coil on the axis, where the magnetic induction will be $\frac{1}{8}$ th to its value at the centre of the coil is
Options:
Solution:
1321 Upvotes
Verified Answer
The correct answer is:
$R \sqrt{3}$
Here the ratio, $\frac{\mathrm{B}_{\mathrm{Centre}}}{\mathrm{B}_{\mathrm{axis}}}=\left(1+\frac{\mathrm{x}^{2}}{\mathrm{R}^{2}}\right)^{3 / 2}$
Also, $B_{\text {axis }}=\frac{1}{8} B_{\text {centre }}$
$\begin{array}{l}
\frac{8}{1}=\left(1+\frac{x^{2}}{R^{2}}\right)^{3 / 2} \Rightarrow 4=1+\frac{x^{2}}{R^{2}} \\
3=\frac{x^{2}}{R^{2}} \Rightarrow X^{2}=3 R^{2} \\
\text { or, } x=\sqrt{3} R
\end{array}$
Also, $B_{\text {axis }}=\frac{1}{8} B_{\text {centre }}$
$\begin{array}{l}
\frac{8}{1}=\left(1+\frac{x^{2}}{R^{2}}\right)^{3 / 2} \Rightarrow 4=1+\frac{x^{2}}{R^{2}} \\
3=\frac{x^{2}}{R^{2}} \Rightarrow X^{2}=3 R^{2} \\
\text { or, } x=\sqrt{3} R
\end{array}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.