Q148 / 160 · TS EAMCET 2024 (09 May Shift 2) · Physics
Two points $A$ and $B$ on the axis of a circular current loop are at distances of 4 cm and $3 \sqrt{3} \mathrm{~cm}$ from the centre of the loop. If the ratio of the induced magnetic fields at points $A$ and $B$ is $216: 125$, the radius of the loop is
- Correct answerA3 cm
- B4 cm
- C5 cm
- D6 cm
Solution
Magnetic field at a point on the axis of circular loop is given by
$\begin{aligned}
& B=\frac{\mu_0}{4 \pi} \frac{2 \pi \mathrm{Nir}^2}{\left(\mathrm{x}^2+\mathrm{r}^2\right)^{3 / 2}} \\
& \therefore \mathrm{~B} \alpha \frac{\mathrm{r}^2}{\left(\mathrm{x}^2+\mathrm{r}^2\right)^{3 / 2}}
\end{aligned}$
Here, $r=$ radius of circular loop $\mathbf{x}=$ distance of a point from the centre of circular loop $\begin{aligned} & \therefore \frac{\mathrm{B}_1}{\mathrm{~B}_2}=\frac{\mathrm{r}^2}{\left[\mathrm{r}^2+(4)^2\right]^{3 / 2}} \times \frac{\left[\mathrm{r}^2+(3 \sqrt{3})^2\right]^{3 / 2}}{\mathrm{r}^2} \\ & \because \frac{\mathrm{~B}_1}{\mathrm{~B}_2}=\frac{216}{125} \text { given } \\ & \therefore \frac{216}{125}=\frac{\left[\mathrm{r}^2+27\right]^{3 / 2}}{\left[\mathrm{r}^2+16\right]^{3 / 2}} \Rightarrow \frac{6}{5}=\frac{\left(\mathrm{r}^2+27\right)^{1 / 2}}{\left(\mathrm{r}^2+16\right)^{1 / 2}} \end{aligned}$
On solving we get, $\mathbf{r}=3$
Here, $r=$ radius of circular loop $\mathbf{x}=$ distance of a point from the centre of circular loop $\begin{aligned} & \therefore \frac{\mathrm{B}_1}{\mathrm{~B}_2}=\frac{\mathrm{r}^2}{\left[\mathrm{r}^2+(4)^2\right]^{3 / 2}} \times \frac{\left[\mathrm{r}^2+(3 \sqrt{3})^2\right]^{3 / 2}}{\mathrm{r}^2} \\ & \because \frac{\mathrm{~B}_1}{\mathrm{~B}_2}=\frac{216}{125} \text { given } \\ & \therefore \frac{216}{125}=\frac{\left[\mathrm{r}^2+27\right]^{3 / 2}}{\left[\mathrm{r}^2+16\right]^{3 / 2}} \Rightarrow \frac{6}{5}=\frac{\left(\mathrm{r}^2+27\right)^{1 / 2}}{\left(\mathrm{r}^2+16\right)^{1 / 2}} \end{aligned}$
On solving we get, $\mathbf{r}=3$
Hence, the correct answer is option: 'A'.