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 field at the centres is the same.
(A) $4$ times of first coil
(B) $6$ times of first coil
(C) $3$ times of first coil
(D) $2$ times of first coil

Answer 1

Hint : Use the formula for magnetic field at the centre of a coil by Biot- Savart law and equate the magnetic fields to find the relationship between the currents. Then use ohm’s law to deduce the voltage relation. The magnetic field of a circular at the centre is given by, $B = \dfrac{{{\mu _0}i}}{{2r}}$ where, $r$ is the radius of the coil , ${\mu _0}$ is the absolute magnetic permeability , $i$ is the current flowing through it.

Complete Step By Step Answer:
we know that the magnetic field at the centre of a coil by Biot –Savart law is $B = \dfrac{{{\mu _0}i}}{{2r}}$ ,where, $r$ is the radius of the coil , ${\mu _0}$ is the absolute magnetic permeability , $i$ is the current flowing through it.
We have given here two coils with the radius of the first coil twice of that second.
So, magnetic field at the centre field is, ${B_1} = \dfrac{{{\mu _0}{i_1}}}{{2{r_1}}}$ and of the second is, ${B_2} = \dfrac{{{\mu _0}{i_2}}}{{2{r_2}}}$
This magnetic field for both the coils is the same. Hence we can write, ${B_1} = {B_2}$
Therefore, $\dfrac{{{\mu _0}{i_1}}}{{2{r_1}}} = \dfrac{{{\mu _0}{i_2}}}{{2{r_2}}}$
Or, $\dfrac{{{i_1}}}{{{r_1}}} = \dfrac{{{i_2}}}{{{r_2}}}$
Now, we have, ${r_1} = 2{r_2}$
Putting the value in the equation we get,
$\dfrac{{{i_1}}}{{2{r_2}}} = \dfrac{{{i_2}}}{{{r_2}}}$
Or, ${i_1} = 2{i_2}$
Now, we know, resistance of a wire of length $l$ with cross section $A$ of resistivity $\rho$ is given by,
$R = \dfrac{{\rho l}}{A}$ . Here we have coils made of the same materials. Hence, resistivity and cross section is the same. So, Resistance of the first coil is, ${R_1} = \dfrac{{\rho {l_1}}}{A}$ . Now, length of the first coil ${l_1} = 2\pi {r_1} = 2\pi (2{r_2})$
So, ${R_1} = \dfrac{{\rho 2\pi (2{r_2})}}{A}$
Resistance of the second coil is, ${R_2} = \dfrac{{\rho {l_2}}}{A}$ . length of the second coil ${l_1} = 2\pi {r_2}$ .
So, ${R_2} = \dfrac{{\rho 2\pi {r_2}}}{A}$
Therefore, by Ohm's law we can find the voltage across the first coil, ${V_1} = {i_1}{R_1}$ . Putting the values,
${V_1} = 2{i_2}\dfrac{{\rho 2\pi (2{r_2})}}{A}$ .
voltage across secondary coil, ${V_2} = {i_2}{R_2}$ . Putting the values, ${V_2} = {i_2}\dfrac{{\rho 2\pi {r_2}}}{A}$
Hence, ratios of their voltages is, $\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{2{i_2}\dfrac{{\rho 2\pi (2{r_2})}}{A}}}{{{i_2}\dfrac{{\rho 2\pi {r_2}}}{A}}}$
Or, $\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{2{i_2}\rho 2\pi (2{r_2})}}{{{i_2}\rho 2\pi {r_2}}}$
Or, $\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{2{i_2}(2{r_2})}}{{{i_2}{r_2}}}$
So, it becomes,
$\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{4}{1}$
Therefore, ${V_1} = 4{V_2}$ .
So, we have to apply a voltage across the first coil that should be four times of the second coil for the magnetic field to be the same.
Hence, option (A) is correct.

Note :
Magnetic field due to any circular coil at its centre depends on the current flowing through it and the radius of the coil. The more the current the more the intensity of the magnetic coil is, less the radius more the magnetic field at the centre of the coil.
Helmholtz coils is an arrangement of similar coils of the same radius to produce a homogeneous magnetic field . Difference is only that we calculate the magnetic field on the axis of it to get a homogenous magnetic field along that region.