Question

Two circular coils 1 and 2 are made from the same wire but the radius of 1st coil is twice that of the 2nd coil. What potential difference in volts should be applied across them so that the magnetic magnetic field at the centres is the same

• A. 4 times of first coil
• B. 6 times of first coil
• C. 2 times of first coil
• D. 3 times of first coil

Answer 1

The correct option is A 4 times of first coil

Magnetic feild due to first coil and second coil is same at center.

Then magnetic field $B$ at center is,

$B=\frac{{\mu }_0{I}_1}{2\left(2r\right)}=\frac{{\mu }_0{I}_2}{2\left(r\right)}=\frac{I_1}{I_2}=2$ ..(1)

As we know that Resistance of coil is related as,

$R=\rho \frac{l}{A}$ where $\rho$=resistivity ,$l$=length ,$A$=area of cross section.

$\rho$ and $A$ is same for both coil but $l_1=2\pi \left(2r\right)$ and $l_2=2\pi \left(r\right)$

If $V_1$ and $V_2$ applied across first and second coil then,

$I_1=\frac{V_1}{R_1}$ and $I_2=\frac{V_2}{R_2}$

$I_1=\frac{V_1}{\rho \frac{l_1}{A}}$..(2) and $I_2=\frac{V_2}{\rho \frac{l_2}{A}}$ ...(3)

From (1),(2),(3),

$\frac{V_1}{l_1}\times \frac{l_2}{V_2}=2$

$\frac{V_1}{V_1}\times \frac{l_2}{l_1}=2$

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

$\frac{V_1}{V_2}=4$

$V_1=4V_2$