Question

A length of wire carries a steady current. It is bent first to form a circular plane of coil of one turn. The same length is now bent more sharply to give a double loop of smaller radius. The magnetic field at the centre caused by the same current is
(A) a quarter of its first value
(B) unaltered
(C) four times of its first value
(D) half of its first value

Answer 1

Hint
The magnetic field in a current carrying loop depends directly on the current flowing through it and inversely on the tightness of the coil. Consequently, a change in radius of the loop will cause the magnetic field to be affected.
The magnetic field at the centre of a current carrying loop is given by
$\Rightarrow B = \dfrac{{{\mu _0}nI}}{{2R}}$ where n is the number of loops, I is the current, and R is the radius of the loop.

Complete step by step answer
We are asked to compare the magnetic field generated at the centre of a loop due to the current flowing through it for two wires of the same length but wound differently.
We know that the magnetic field is given as:
$\Rightarrow B = \dfrac{{{\mu _0}nI}}{{2R}}$
For the initial case, let us assume a wire of length l and circular loop of radius R. This implies the total length will be equal to the circumference of the loop as:
$\Rightarrow l = 2\pi R$
$\Rightarrow R = \dfrac{l}{{2\pi }}$
This gives us the magnetic field for one loop as:
$\Rightarrow B = \dfrac{{{\mu _0} \times 1 \times I}}{{2\left( {\dfrac{l}{{2\pi }}} \right)}} = \dfrac{{{\mu _0}\pi I}}{l}$ [Eq. 1]
For the second case, the same length gives a double loop. For a radius of r, this length will be equal to 2 times the circumference as:
$\Rightarrow l = 2 \times 2\pi r$
$\Rightarrow r = \dfrac{l}{{4\pi }}$
Also, since there are two loops now, n will be equal to 2. Putting the values to find new magnetic field:
$\Rightarrow {B_{new}} = \dfrac{{{\mu _0} \times 2 \times I}}{{2\left( {\dfrac{l}{{4\pi }}} \right)}} = \dfrac{{{\mu _0}4\pi I}}{l}$ [Eq. 2]
Dividing Eq.1 by Eq. 2 gives us:
$\Rightarrow \dfrac{B}{{{B_{new}}}} = \dfrac{{\dfrac{{{\mu _0}\pi I}}{l}}}{{\dfrac{{{\mu _0}4\pi I}}{l}}} = \dfrac{1}{4}$
This implies ${B_{new}} = 4B$, and hence the correct answer is option (C).

Note
We know that the current flowing through a wire tends to produce a magnetic field around it. For a circular loop, each small segment of the wire generates a magnetic field of its own. This causes the magnetic field lines to be concentrated densely at the centre, and hence the magnetic field intensity is the strongest at the centre of the loop.