Question

A circular coil of radius 2.0 cm carries a current of 1.0 A. If the coil has 100 turns, find the intensity of magnetic field at the centre of the coil.

Hint:

Apply \( B=\dfrac{\mu_0 N I}{2r} \); remember to convert the radius from cm to metres.

Answer 1

Step 1: Field of one turn first.
The magnetic field at the centre of a single circular loop is \(B_1 = \dfrac{\mu_0 I}{2r}\). A coil of \(N\) closely wound turns produces \(N\) times this value, so \(B = N B_1 = \dfrac{\mu_0 N I}{2r}\).

Step 2: Convert units.
\(r = 2.0\ \text{cm} = 0.02\ \text{m}\), \(I = 1.0\ \text{A}\), \(N = 100\).

Step 3: Compute the single-turn field.
\(B_1 = \dfrac{(4\pi\times10^{-7})(1.0)}{2(0.02)} = \dfrac{4\pi\times10^{-7}}{0.04} = \pi\times10^{-5}\ \text{T}.\)

Step 4: Multiply by number of turns.
\(B = 100 \times \pi\times10^{-5} = \pi\times10^{-3}\ \text{T}.\)

Step 5: Numerical answer.
Taking \(\pi = 3.14\),
\[\boxed{B \approx 3.14\times10^{-3}\ \text{T}\ (3.14\ \text{mT})}\]
directed perpendicular to the plane of the coil along its axis.