A closely wound circular coil of radius 5 cm produces a magnetic field of $$37.68 \times 10^{-4}$$ T at its center. The current through the coil is _____ A. [Given, number of turns in the coil is 100 and $$\pi = 3.14$$]

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

$$ B = \frac{\mu_0 N I}{2R} $$

Here, $$ B $$ is the magnetic field, $$ \mu_0 $$ is the permeability of free space ($$ 4\pi \times 10^{-7} $$ T m/A), $$ N $$ is the number of turns, $$ I $$ is the current, and $$ R $$ is the radius of the coil.

We are given:

• Radius $$ R = 5 $$ cm = 0.05 m (converted to meters for SI units),
• Magnetic field $$ B = 37.68 \times 10^{-4} $$ T,
• Number of turns $$ N = 100 $$,
• $$ \pi = 3.14 $$.

First, express $$ \mu_0 $$ using the given $$ \pi $$:

$$ \mu_0 = 4 \pi \times 10^{-7} = 4 \times 3.14 \times 10^{-7} = 12.56 \times 10^{-7} = 1.256 \times 10^{-6} \text{ T m/A} $$

Now, rearrange the formula to solve for $$ I $$:

$$ I = \frac{2 R B}{\mu_0 N} $$

Substitute the known values:

$$ I = \frac{2 \times 0.05 \times 37.68 \times 10^{-4}}{1.256 \times 10^{-6} \times 100} $$

Compute the numerator:

$$ 2 \times 0.05 = 0.1 $$

$$ 0.1 \times 37.68 \times 10^{-4} = 0.1 \times 0.003768 = 0.0003768 = 3.768 \times 10^{-4} $$

Compute the denominator:

$$ 1.256 \times 10^{-6} \times 100 = 1.256 \times 10^{-4} $$

Now, divide the numerator by the denominator:

$$ I = \frac{3.768 \times 10^{-4}}{1.256 \times 10^{-4}} = \frac{3.768}{1.256} $$

Perform the division:

$$ 3.768 \div 1.256 = 3 $$

Therefore, the current $$ I $$ is 3 A.

Hence, the correct answer is 3.