A tightly wound 100 turn coil of radius 5 cm is carrying a certain amount of current. The magnitude of the magnetic field at the center of the coil is $2π × 10^{-4}T$. The value of current flowing through the coil is

0.5 A

The correct answer is Option (4) → 0.5 A

Given:

Number of turns: $N = 100$

Radius of coil: $R = 5~\text{cm} = 0.05~\text{m}$

Magnetic field at center: $B = 2\pi \times 10^{-4}~\text{T}$

Magnetic field at the center of a circular coil:

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

Where $\mu_0 = 4\pi \times 10^{-7}~\text{T·m/A}$

Solving for current $I$:

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

Substitute values:

$I = \frac{2 \cdot (2 \pi \times 10^{-4}) \cdot 0.05}{4 \pi \times 10^{-7} \cdot 100}$

$I = \frac{2 \cdot 2 \pi \cdot 10^{-4} \cdot 0.05}{4 \pi \cdot 10^{-5}}$

$I = \frac{2 \cdot 2 \cdot 0.05 \cdot 10^{-4}}{4 \cdot 10^{-5}} = \frac{0.2 \cdot 10^{-4}}{4 \cdot 10^{-5}} = \frac{2 \cdot 10^{-5}}{4 \cdot 10^{-5}} = 0.5~\text{A}$

Answer: $I = 0.5~\text{A}$