Two circular coils 1 and 2 are made from the same wire. The radius of the first coil is twice that of the…
Two circular coils 1 and 2 are made from the same wire. The radius of the first coil is twice that of the second coil. What is the ratio of potential difference applied across them $V_1 / V_2$, so that the magnetic field at their centre is the same?
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Solution
Magnetic field due to first coil and second coil is same at centre.
Then, magnetic field at centre is
$
B=\frac{\mu_0 I_1}{2(2 r)}=\frac{\mu_0 I_2}{2(r)} \Rightarrow \frac{I_1}{I_2}=2...(i)
$
As we know that, resistance of the coil is related as
$
R=\rho \frac{l}{A}
$
where, $\rho=$ resistivity
$l=$ length
$A=$ area of cross-section
$\rho$ and $A$ are same for both coils but
$l_1=2 \pi(2 r)$ and $l_2=2 \pi(r)$.
$\Rightarrow \quad \frac{l_2}{l_1}=\frac{1}{2}...(ii)$
If $V_1$ and $V_2$ are potential difference applied across first and second coil, then
$
\text { and } \quad \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{l_1}{A}}...(iii) \\
I_2 & =\frac{V_2}{\rho \frac{l_2}{A}}...(iv)
\end{aligned}
$
From Eqs. (i), (iii) and (iv), we get
$
\begin{array}{clrl}
& & & V_1 / l_1 \times l_2 / V_2=2 \\
\Rightarrow & & \frac{V_1}{V_2} \times \frac{l_2}{l_1}=2 \\
\Rightarrow & & \frac{V_1}{V_2} \times \frac{1}{2}=2 \\
\Rightarrow & & \frac{V_1}{V_2}=4
\end{array}
$
[Form Eq. (ii) ]