Question

A 100-turn closely wound circular coil of radius 5 cm has a magnetic field of 3.14 $\times$ 10⁻³ T at its centre. The current flowing through the coil, and the magnitude of the magnetic moment of this coil are, respectively: (Take $\mu_0 = 4\pi \times 10^{-7}$ T m/A)

• A. 2 A, 4 A m²
• B. 2.5 A, 20 A m²
• C. 2.5 A, 2 A m²
• D. 2 A, 10 A m²

Hint:

In competitive exams, $3.14$ is often used interchangeably with $\pi$. Canceling $\pi$ on both sides of your equations early on usually makes the calculation much simpler.

Answer 1

The Correct Option is C

Step 1: Understanding the Topic:
This problem is from "Magnetic Effects of Current." It covers two fundamental concepts: the magnetic field produced by a circular current loop at its center and the definition of magnetic dipole moment for a coil. A current-carrying coil acts as a magnet, and these equations quantify its strength.
Step 2: Key Formulas and Approach:

Magnetic field at the center of a coil: $B = \frac{\mu_0 N I}{2r}$.
Magnetic moment of a coil: $M = N \cdot I \cdot A$.
Area of a circular coil: $A = \pi r^2$.

Step 3: Detailed Explanation:

Identify given values: $N = 100$, $r = 5 \text{ cm} = 0.05 \text{ m}$, and $B = 3.14 \times 10^{-3} \text{ T}$ (Note that $3.14 \approx \pi$).
Find Current ($I$): Rearrange the field formula: \[ I = \frac{2 \cdot r \cdot B}{\mu_0 \cdot N} \] \[ I = \frac{2 \times 0.05 \times (\pi \times 10^{-3})}{4\pi \times 10^{-7} \times 100} \]
The $\pi$ cancels out: \[ I = \frac{0.1 \times 10^{-3}}{4 \times 10^{-5}} = \frac{10^{-4}}{4 \times 10^{-5}} = \frac{10}{4} = 2.5 \text{ A} \]
Find Magnetic Moment ($M$):
First, calculate Area: $A = \pi \times (0.05)^2 = 3.14 \times 0.0025 = 0.00785 \text{ m}^2$.
Then, $M = 100 \times 2.5 \times 0.00785 = 250 \times 0.00785$.
$M = 1.9625 \text{ A m}^2 \approx 2 \text{ A m}^2$.

Thus, the current is 2.5 A and the moment is 2 units.
Step 4: Final Answer:
The current flowing is 2.5 A and the magnetic moment is 2 A m².