Question

When current is passed through a circular wire prepared from a long conducting wire, the magnetic field produced at its centre is $B$ . Now a loop having two turns prepared from the same wire and the same current is passed through it. The magnetic field at its centre will be
(A) $4B$
(B) $\dfrac{B}{4}$
(C) $\dfrac{B}{2}$
(D) $16B$

Answer 1

The magnetic field is directly proportional to the number of turns of the wire. As the number of turns is increased for the same wire, the radius of the loop will decrease. The magnetic field is inversely proportional to the radius of the loop.

Formula used: In this solution we will be using the following formula;
$B = \dfrac{{\mu NI}}{{2r}}$ where $B$ is the magnetic field, $\mu$ is the permeability of free space, $I$ is the current flowing through the wire, and $r$ is the distance of the point of interest from the wire.

Complete step by step solution:
Generally, when current flows through a conductor, a magnetic field is created around the conductor. The magnetic field in general, is given by the Biot-savart law which can be written in its constant form as
$B = \dfrac{{\mu NI}}{{2r}}$ where $B$ is the magnetic field, $\mu$ is the permeability of free space, $I$ is the current flowing through the wire, and $r$ is the distance of the point of interest from the wire (radius for the case of the centre of a circular loop).
Hence, since the current and permeability are constant, we can write that
$B = k\dfrac{N}{r}$ .
Hence for the first situation, ${B_1} = k\dfrac{{{N_1}}}{{{r_1}}}$ , and for the second, ${B_2} = k\dfrac{{{N_2}}}{{{r_2}}}$
Now, ${N_2} = 2{N_1}$ and ${r_2} = \dfrac{{{r_1}}}{2}$
Then ${B_2} = k\dfrac{{2{N_1}}}{{\dfrac{{{r_1}}}{2}}} = k\dfrac{{2{N_1}}}{{{r_1}}} \times 2$
${B_2} = 4k\dfrac{{{N_1}}}{{{r_1}}}$ .
Now, since ${B_1} = k\dfrac{{{N_1}}}{{{r_1}}}$ , by replacing into the above equation, we see that
${B_2} = 4{B_1}$
The magnetic field ${B_1}$ according to the question is ${B_1} = B$
Then, ${B_2} = 4B$
Hence, the correct answer is option A.

Note:
For clarity, it can be proven that the radius of the second loop is half that first loop as follows:
Say the first wire as a length of $l$ , then
$l = 2\pi {r_1}$
In the second situation, there are two loops with the same length, then
$l = 2\left( {2\pi {r_2}} \right)$
Hence, we can observe that the length would be equal, if we allow ${r_2} = \dfrac{{{r_1}}}{2}$
Hence, it is half the original radius.