Question

A coil of one turn is made of a wire of certain length and then from the same length a coil of two turns is made. If the same current is passed in both the cases, then the ratio of the magnetic induction at their centres will be:
A. 2:1
B. 1:4
C. 4:1
D. 1:2

Answer 1

A good starting point would be to determine the expression for magnetic induction at the centre of a circular coil and determine the factors it depends on and their proportionality. In other words, since the same current is passed through in both cases and the wire is always of the same length, arrive at the relation between the magnetic field applied and the radius of the coil and the number of turns. Note that as the radius decreases, the intensity of magnetic induction increases. Think of how lesser radius and number of turns is related and you should be able to arrive at the correct answer.

Formula used:
Magnetic induction at the centre of a current carrying coil $B = \dfrac{\mu_0 nI}{2r}$, where $\mu_0$ is the magnetic permeability, n is the number of turns in the coil, I is the current passing through the coil, and r is the radius of the coil.

Complete answer:
We can define electromagnetic induction as the phenomenon in which electric current is induced by appropriately varying magnetic fields.
As for a current carrying circular coil, the magnetic induction at the centre of a current carrying coil is given as
$B = \dfrac{\mu_0 nI}{2r} \Rightarrow B \propto \dfrac{n}{r}$
Let us begin with the initial case, where a coil of $n_1 =1$ turn is made from a certain length. Let the radius of this coil be $r_1$
The length of the coil can be given as the perimeter of the circle formed by a turn in the coil times the number of turns, i.e.,
$L = 2\pi r_1 \times n_1 \Rightarrow r_1 = \dfrac{L}{2\pi n_1} = \dfrac{L}{2\pi}$
The magnetic induction is given as:
$B_1 \propto \dfrac{1}{r_1}$
Now, in the second case, we have the radius of the coil to be $r_2$ made up of $n_2 = 2$ turns made from the same length.
$L = 2\pi r_2 \times n_2 \Rightarrow r_2 = \dfrac{L}{2\pi n_2} = \dfrac{1}{2}\dfrac{L}{2\pi} = \dfrac{1}{2} r_1$
The magnetic induction is given as:
$B_2 \propto \dfrac{2}{r_2} = \dfrac{2}{\dfrac{r_1}{2}} = \dfrac{4}{r_1}$
Therefore,
$\dfrac{B_1}{B_2} = \dfrac{1}{r_1}\times \dfrac{r_1}{4} = \dfrac{1}{4}$
Therefore the ratio of the magnetic induction $B_1$:$B_2$ = 1:4

So, the correct answer is “Option B”.

Note:
Remember that a coil, in a geometric sense, is just multiple circular loops put together. This is why we defined the length of the coil as we did in our evaluation.
Do not forget that at the centre of a circular current carrying loop, the magnetic field lines are straight and are produced in the same direction within the loop. Subsequently, the direction of the magnetic field at the centre of a circular coil is along the axis of the coil.