Question Details

A tightly wound 100 turns coil of radius 10 cm carries a current of 7 A. The magnitude of the magnetic field at the centre of the coil is (Take permeability of free space as 4π × 10–7 SI units) :

Options

A

44 mT

B

4.4 T

C

4.4 mT

D

44 T

Show Answer

Correct Answer :

Option C

4.4 mT

4.4 mT

Solution :

**Step 1 – Identify the formula**
For a circular coil of radius R carrying N turns of current I, the magnetic field at the centre is

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

where μ₀ = 4π × 10⁻⁷ T·m/A (permeability of free space).

**Step 2 – Convert all quantities to SI units**
Radius R = 10 cm = 0.10 m
Number of turns N = 100
Current I = 7 A

**Step 3 – Plug the numbers into the formula**

B = \frac{(4\pi \times 10^{-7}\,\text{T·m/A}) \times 100 \times 7\,\text{A}}{2 \times 0.10\,\text{m}}

**Step 4 – Simplify the expression**

\frac{100 \times 7}{2 \times 0.10} = \frac{700}{0.20} = 3500

Hence

B = (4\pi \times 10^{-7}) \times 3500

**Step 5 – Perform the multiplication**

4 \times 3500 = 14000

So

B = 14000\pi \times 10^{-7}\,\text{T}

Convert the power of ten:

14000 \times 10^{-7} = 1.4 \times 10^{-3}

Therefore

B = 1.4 \times 10^{-3}\,\pi\ \text{T}

**Step 6 – Approximate π (≈ 3.1416)**

B \approx 1.4 \times 10^{-3} \times 3.1416\ \text{T} \approx 4.398 \times 10^{-3}\ \text{T}

**Step 7 – Express in millitesla**

4.398 \times 10^{-3}\ \text{T} = 4.398\ \text{mT} \approx 4.4\ \text{mT}

**Conclusion**
The magnetic field at the centre of the coil is **4.4 mT**, which matches the given correct option.

Unlock Our Free Library

Access expert-curated educational resources and study materials—completely free.

Ask AI Tutor
5 left
Q1 View Question & Options
AI Tutor is solving this question...
Reading question context & options...