Two circular coils 1 and 2 are made from the same wire but the radius of the first coil is twice that of the…
Two circular coils 1 and 2 are made from the same wire but the radius of the first coil is twice that of the second coil. What is the ratio of potential difference applied across them, so that the magnetic field at their centres is same?
$2: 3$
$6: 4$
$3: 2$
$4: 1$
Solution
Magnetic feild due to first coil and second coil is same at center.
Then magnetic field $B$ at center is,
$\mathrm{B}=\frac{\mu_0 \mathrm{I}_1}{2(2 \mathrm{r})}=\frac{\mu_0 \mathrm{I}_2}{2(\mathrm{r})}=\frac{\mathrm{I}_1}{\mathrm{I}_2}=2$ ...(1)
As we know that Resistance of coil is related as, $\mathrm{R}=\rho \frac{\mathrm{l}}{\mathrm{A}}$ where $\rho=$ resistivity, $\mathrm{l}=$ length, $\mathrm{A}=$ area of cross section.
$\rho$ and $A$ is same for both coil but $l_1=2 \pi(2 r)$ and $l_2=2 \pi(r)$
If $V_1$ and $V_2$ applied across first and second coil then,
$\begin{aligned}
& I_1=\frac{V_1}{R_1} \text { and } I_2=\frac{V_2}{R_2} \\
& I_1=\frac{V_1}{\rho \frac{\mathrm{l}_1}{A}} . .(2) \text { and } \mathrm{I}_2=\frac{\mathrm{V}_2}{\rho \frac{\mathrm{l}_2}{A}} \ldots \text { (3) }
\end{aligned}$
From (1),(2),(3),
$\frac{V_1}{l_1} \times \frac{l_2}{V_2}=2$
$\frac{\mathrm{V}_1}{\mathrm{~V}_1} \times \frac{\mathrm{l}_2}{\mathrm{l}_1}=2$
$\frac{\mathrm{V}_1}{\mathrm{~V}_2} \times \frac{1}{2}=2$
$\begin{aligned}
& \frac{\mathrm{V}_1}{\mathrm{~V}_2}=4 \\
& \mathrm{~V}_1=4 \mathrm{~V}_2
\end{aligned}$