Question

Two circular coils P and Q are made from similar wire, but radius of Q is twice that of P. What should be the value of potential difference across them so that the magnetic induction at their centre may be same?

• A. $V_q=2V_p$
• B. $V_q=3V_p$
• C. $V_q=4V_p$
• D. $V_q=\frac{1}{4}{V}_p$

Answer 1

The correct option is C $V_q=4V_p$

Magnet field induction at the centre of the coil $B=\frac{{\mu }_{o}i}{2r}$

where $i$ and $r$ represents the current flowing and radius of that coil respectively.

Resistance of the circular coil $R=k\left(2\pi r\right)$ where $k$ is a constant

$\therefore$ Potential difference across the coil $V=iR=i\left(2\pi r\right)k$

$\therefore$ $B=\frac{{\mu }_o}{2}\times \frac{V/2\pi rk}{r}=\frac{{\mu }_o}{4\pi }\frac{V}{r^{2}k}$

$⟹$ $V=\frac{4\pi B}{{\mu }_o}{r}^{2}k$ OR $V\propto r^2$ ($\because B$ is same for two loops)

$\therefore$ $\frac{V_Q}{V_p}=\frac{r_Q^2}{r_p^2}=\frac{(2r_p{)}^2}{r_p^2}$

$⟹$ $V_Q=4V_p$