A length of wire carries a steady current, is bent first to form a plane circular coil of one turn, same length now bent more sharply to give three turns of smaller radius. Magnetic field becomes at centre:A) 3 timesB) 1/3 timesC) 9 timesD) Unchanged

The magnetic field at the center of a circular coil of wire is given by the formula:

B = (μ₀ * N * I) / (2 * R)

Where:
B is the magnetic field strength at the center.
μ₀ is the permeability of free space (a constant).
N is the number of turns of the coil.
I is the current flowing through the wire.
R is the radius of the coil.

Let's compare the initial coil (one turn) to the final coil (three turns of smaller radius):

Initial coil:

Number of turns (N₁) = 1
Radius (R₁)
Final coil:

Number of turns (N₂) = 3
Radius (R₂) = Smaller than R₁
Now, let's compare the magnetic fields at the center for the two cases:

For the initial coil (one turn):
B₁ = (μ₀ * 1 * I) / (2 * R₁)

For the final coil (three turns of smaller radius):
B₂ = (μ₀ * 3 * I) / (2 * R₂)

Now, we need to determine the relationship between B₁ and B₂:

B₂ / B₁ = [(μ₀ * 3 * I) / (2 * R₂)] / [(μ₀ * 1 * I) / (2 * R₁)]

B₂ / B₁ = (3 * I) / (1 * I) * (R₁ / R₂)

B₂ / B₁ = 3 * (R₁ / R₂)

Since the wire is bent more sharply to form three turns of smaller radius (R₂ < R₁), we can say that R₁ / R₂ > 1, which means that B₂ / B₁ > 3.

So, the magnetic field at the center of the final coil with three turns of smaller radius is greater than three times the magnetic field at the center of the initial coil. Therefore, the correct answer is:

A) 3 times