Question

A circular coil of one turn carries a current I. The same wire is then bent to form a small circular coil of 2 turns and the same current is passed through it. What is the relation between the fields at the centre of the coils in the second and first case?
A. $B_2 = B_1$
B. $B_2 = 2B_1$
C. $B_2 = \dfrac{B_1}{2}$
D. $B_2 = 4B_1$

Answer 1

Hint: Ampere's law states that line integral of magnetic field taken along a closed loop is equal to $\mu_0$ times the current and enclosed in the loop. Therefore, we can draw an imaginary loop of any radius around a conductor and find the magnitude of the magnetic field produced by it.

Formula used:
The magnetic field at the centre of a current carrying circular coil is.
$B = \dfrac{\mu_0 2NI}{r}$
Where N is the number of turns and r is the radius of the coil.

Complete answer:
We assume that the radius of a single turn coil is R and the radius of a coil with two turns is r. The relation between the two can be obtained as:

$2 \pi R = 2 \times (2 \pi r)$
R = 2r

If we consider the coil with single turn and draw a closed loop of radius R, passing through the centre around it, then with the help of ampere's law we can write:
$B_1 \times 2 \pi R = \mu_0 I$
Or,
$B_1 = \dfrac{\mu_0 I}{2 \pi R}$

Now for the case of small coil with two turns, we write NI in place of I (or 2I in our case) and substitute the R in the above formula, we get:
$B_2 = \dfrac{\mu_0 4 I}{2 \pi r}$
Substituting the value of $B_1$ obtained previously into this gives us:
$B_2 = 4 B_1$

Additional Information:
The magnetic field at the centre of the circular loop can also be found by the use of Biot-Savart law. The angle between the current element and r will be a right angle always. The 'formula used' section here contains the result from that derivation. The difference between that formula and the one obtained using ampere's law is that Biot-Savart produces the total magnetic field due to the whole coil while ampere's law is a good alternative to it for the sake of comparison in our case.

Note:
We can also draw a circular loop of radius r for the case of a smaller coil and derive the required expression $B_2$ by using ampere's law. In this case, the current enclosed will be twice of I because there are two turns.