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) :
Correct Answer :
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
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**
**Step 4 – Simplify the expression**
Hence
**Step 5 – Perform the multiplication**
So
Convert the power of ten:
Therefore
**Step 6 – Approximate π (≈ 3.1416)**
**Step 7 – Express in millitesla**
**Conclusion**
The magnetic field at the centre of the coil is **4.4 mT**, which matches the given correct option.
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