Question

Magnetic fields at two points on the axis of a circular coil at a distance of$0.05m$ and$0.2m$ from the center are in the ratio$8:1$. The radius of coil is ______

• A. $0.15m$
• B. $0.2m$
• C. $0.1m$
• D. $1.0m$

Answer 1

The correct option is C

$0.1m$

Step 1. Given

The distance of the first point$=0.05m$

The distance of the first point$=0.2m$

The ratio of the magnetic fields$=8:1$

Step 2. Formula used:

The magnetic field at a point on the axis of a circular coil of radius $r$ at $x$ distance away from the center of the coil is,

$B=\frac{{\mu }_{0}Ni{r}^2}{2{(r^2+x^2)}^{{\displaystyle \frac{3}{2}}}}$ ,($i=$Current flowing through the coil, $N=$Number of turns)

Step 3. Calculating the radius of the coil:

Consider the following figure:

$$\begin{gathered} \Rightarrow \frac{B_1}{B_2}=\frac{{\mu }_{0}Ni{r}^2}{2{(r^2+0.{05}^2)}^{{\displaystyle \frac{3}{2}}}}\times \frac{2{(r^2+0.2^2)}^{\frac{3}{2}}}{{\mu }_{0}Ni{r}^2}=\frac{8}{1} \\ \Rightarrow \frac{{(r^2+0.2^2)}^{\frac{3}{2}}}{{(r^2+0.{05}^2)}^{\frac{3}{2}}}=\frac{8}{1}=2^3 \\ \Rightarrow \frac{(r^2+0.2^2)}{(r^2+0.{05}^2)}=2^2=4 \\ \Rightarrow 4r^2+0.01=r^2+0.04 \\ \Rightarrow r^2=\frac{0.04-0.01}{3} \\ \Rightarrow r=0.1m \end{gathered}$$

Thus, the radius is $0.1m$.

Hence, option C is the correct answer.