Question

Two circular coils $1$ and $2$ are made from the same wire but the radius of the 1st coil is twice that of the $2nd$ coil. What is the ratio of potential difference applied across them, so that the magnetic field at their centers is the same?

• A. $3$
• B. $1$
• C. $6$
• D. $2$

Answer 1

Answer: (B)

$1$

Given :

$r_1=2r_2$

$B_1=B_2$

To find :

$\frac{V_1}{V_2}$

Solution :

We know that B at center of circular coil is given by: $B=\frac{{\mu }_{0}i}{2r}$

Form the given statement

$\frac{{\mu }_0{i}_1}{2r_1}=\frac{{\mu }_0{i}_2}{2r_2}$

$\therefore \frac{i_1}{i_2}=\frac{r_1}{r_2}$

$\therefore \frac{i_1}{i_2}=\frac{2r_2}{r_2}$

$\therefore \frac{i_1}{i_2}=2$

Also, we know that $\frac{V_1}{V_2}=\frac{i_1{R}_1}{i_2{R}_2}$

Now, $\frac{R_1}{R_2}=\frac{r_2}{r_1}$

Hence , $\frac{V_1}{V_2}=\frac{i_1{r}_2}{i_2{r}_1}$

$\frac{V_1}{V_2}=\frac{2}{2}=1$