Question

Two circular coils $\text{A}$ and $\text{B}$ of same radius are made from wire of similar materials but the radius of wire $\text{B}$ is twice that of $\text{A}$. The ratio of values of the potential difference across $\text{A}$ and $\text{B}$, such that the magnetic field induction at their centre may be same, will be

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is D $4$

For a wire of radius $r$, resistance $R$ will be

$R\propto \frac{1}{r^2}$ $(\because R\propto \frac{1}{A},A=\pi r^2)$

Also, from Ohm's law

$i=\frac{\Delta V}{R}$

$\therefore i\propto r^2\Delta V..............\left(1\right)$

The magnetic field at centre of circular wire (radius $a$ for circular loop)

$B=\frac{{\mu }_{0}i}{2a}$

$B\propto i\left({\mu }_0\&a\text{are constant}\right)$

From equation $\left(1\right)$

$B\propto r^2\Delta V$

$\Delta V\propto \frac{B}{r^2}$

For $B$ to be same for both sides,

$\frac{(\Delta V{)}_A}{(\Delta V{)}_B}=\frac{r{{}_B}^2}{r{{}_A}^2}={\left(\frac{r_B}{r_A}\right)}^2$

$\frac{\Delta V_A}{\Delta V_B}={\left(\frac{2r_A}{r_A}\right)}^2(\text{Given,}{r}_B=2r_A)$

$\therefore \frac{\Delta V_A}{\Delta V_B}=4$

Hence, option $\left(d\right)$ is the correct choice.