Question

Consider a tightly wound 100 turn coil of radius 10 cm carrying a current of 2A. The magnitude of the magnetic field at the ntre of the coil is

• A. $3.14 \times 10^{-4} \text{ T}$
• B. $6.28 \times 10^{-4} \text{ T}$
• C. $12.56 \times 10^{-4} \text{ T}$
• D. $0$

Hint:

Circular Coil Magnetic Field:

• Single loop: $B = \dfrac\mu_0 I2r$
• For $N$ turns: $B = \dfrac\mu_0 N I2r$
• Always express radius in meters.

Answer 1

The Correct Option is C

Given: $N = 100$, $r = 0.1$ m, $I = 2$ A
\[ B = \frac{\mu_0 N I}{2r} = \frac{4\pi \times 10^{-7} \cdot 100 \cdot 2}{2 \cdot 0.1} = 4\pi \cdot 10^{-4} \text{ T} \approx 12.57 \times 10^{-4} \text{ T} \]

Answer 2

Step 1: Write down the given data
Number of turns, \( N = 100 \)
Radius of the coil, \( r = 10 \text{ cm} = 0.1 \text{ m} \)
Current, \( I = 2 \text{ A} \)

Step 2: Recall the formula for magnetic field at the center of a circular coil
The magnetic field \( B \) at the center of a coil with \( N \) turns carrying current \( I \) is:
\[ B = \frac{\mu_0 N I}{2r} \]
where \( \mu_0 = 4 \pi \times 10^{-7} \, \text{T·m/A} \) is the permeability of free space.

Step 3: Substitute the known values
\[ B = \frac{4 \pi \times 10^{-7} \times 100 \times 2}{2 \times 0.1} = \frac{4 \pi \times 10^{-7} \times 200}{0.2} = \frac{800 \pi \times 10^{-7}}{0.2} \]

Step 4: Simplify the expression
\[ B = 4000 \pi \times 10^{-7} = 4000 \times 3.1416 \times 10^{-7} = 12566.4 \times 10^{-7} = 1.25664 \times 10^{-3} \text{ T} \]

Step 5: Express the answer in given format
\[ B = 12.56 \times 10^{-4} \text{ T} \]

Step 6: Conclusion
The magnitude of the magnetic field at the center of the coil is \( 12.56 \times 10^{-4} \text{ T} \).