Question

Consider two thin identical conducting wires covered with very thin insulating material. One of the wires is bent into a loop and produces magnetic field ${{B}_{1}}$, at its centre when a current $I$ passes through it. The second wire is bent into a coil with three identical loops adjacent to each other produces a magnetic field ${{B}_{2}}$at the centre of the loops when current $\dfrac{I}{3}$ passes through it. The ratio ${{B}_{1}}:{{B}_{2}}$ is
$\begin{aligned} & A)1:1 \\\ & B)1:3 \\\ & C)1:9 \\\ & D)9:1 \\\ \end{aligned}$

Answer 1

Two thin identical conducting wires covered with very thin insulating material suggests that both the wires have the same length. One wire has just one number of turns while the other wire has three numbers of turns. Radius of each coil is determined separately from the circumference of each loop of wire. Finally, the magnetic field at the centre of each coil is determined using these radii, currents flowing through each coil as well as the number of turns of each coil.
Formula used:
$B=\dfrac{{{\mu }_{0}}nI}{2r}$

Complete answer:
We know that magnetic field at the centre of a current carrying coil is given by
$B=\dfrac{{{\mu }_{0}}nI}{2r}$
where
$B$ is the magnetic field at the centre of a current carrying coil of $n$ number of turns
$I$ is the current flowing through the coil
$r$ is the radius of the current carrying coil
${{\mu }_{0}}$ is the magnetic constant
Let this be equation 1.
Coming to our question, we are provided with two thin identical conducting wires covered with very thin insulating material. One of the wires is bent into a loop and produces magnetic field ${{B}_{1}}$, at its centre when a current $I$ passes through it. The second wire is bent into a coil with three identical loops adjacent to each other and produces a magnetic field ${{B}_{2}}$ at the centre of the loops when current $\dfrac{I}{3}$ passes through it. We are supposed to find the ratio ${{B}_{1}}:{{B}_{2}}$.
Firstly, two thin identical conducting wires covered with very thin insulating material suggests that both the wires have the same length. Let us call this equal length of both the wires $L$. Clearly, if radii of both coils are represented as ${{r}_{1}}$ and ${{r}_{2}}$ respectively, then, the circumferences of both current carrying coils are given by
$L=2\pi {{r}_{1}}=3\times 2\pi {{r}_{2}}=6\pi {{r}_{2}}$
where
$L$ is the total length of each coil of wire
${{r}_{1}}$ is the radius of the first coil which has just one number of turn
${{r}_{2}}$ is the radius of the second coil, which has three number of turns
Let this be equation 2.
Now, let us use equation 1 to find the magnetic field at the centre of each coil. Magnetic field at the centre of the first coil ${{B}_{1}}$ is given by
${{B}_{1}}=\dfrac{{{\mu }_{0}}nI}{2{{r}_{1}}}=\dfrac{{{\mu }_{0}}I}{2\left( \dfrac{L}{2\pi } \right)}={{\mu }_{0}}I\times \dfrac{2\pi }{2L}=\dfrac{{{\mu }_{0}}\pi I}{L}$
where
${{B}_{1}}$ is the magnetic field at the centre of the first coil with just one number of turn $(n=1)$
$I$ is the current flowing through the first coil, as given in the question
${{r}_{1}}=\dfrac{L}{2\pi }$, is the radius of the first coil, from equation 2
$L$ is the length of the wire, which is bent into the shape of a coil of one turn
Let this be equation 3.
Similarly, using equation 1, magnetic field at the centre of the second coil ${{B}_{2}}$ is given by
${{B}_{2}}=\dfrac{{{\mu }_{0}}n\left( \dfrac{I}{3} \right)}{2{{r}_{2}}}=\dfrac{{{\mu }_{0}}\times 3\times \left( \dfrac{I}{3} \right)}{2\left( \dfrac{L}{6\pi } \right)}=\dfrac{3{{\mu }_{0}}I}{3}\times \dfrac{6\pi }{2L}=\dfrac{3{{\mu }_{0}}\pi I}{L}$
where
${{B}_{2}}$ is the magnetic field at the centre of the second coil with three number of turns $(n=3)$
$\dfrac{I}{3}$ is the current flowing through the second coil, as given in the question
${{r}_{2}}=\dfrac{L}{6\pi }$, is the radius of the second coil, from equation 2
$L$ is the length of the wire, which is bent into the shape of a coil of three turns
Let this be equation 4.
Now, to determine the ratio ${{B}_{1}}:{{B}_{2}}$, let us divide equation 3 by equation 4, as follows
$\dfrac{{{B}_{1}}}{{{B}_{2}}}=\left( \dfrac{\dfrac{{{\mu }_{0}}\pi I}{L}}{\dfrac{3{{\mu }_{0}}\pi I}{L}} \right)=\dfrac{1}{3}\Rightarrow {{B}_{1}}:{{B}_{2}}=1:3$

Therefore, the correct answer is option $B$.

Note:
Students need to understand that equation 2 is the most important part of the whole solution. If this expression is deduced, the rest of the solution can easily be sorted out. To highlight the same, we are provided with two identical conducting wires, which suggests that both the wires have equal lengths. Length of the first coil is equal to the circumference of the first coil of single turn whereas length of the second coil is thrice the circumference of second coil of three turns. It also needs to be noted that circumference of a coil has the formula $2\pi R$, from which the radius of each coil can be sorted out separately.