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: Write the given data.
Radius \( r = 2.0\ \text{cm} = 2.0\times10^{-2}\ \text{m} \)
Current \( I = 1.0\ \text{A} \)
Number of turns \( N = 100 \)
Permeability of free space \( \mu_0 = 4\pi\times10^{-7}\ \text{T m A}^{-1} \)

Step 2: State the formula.
The magnetic field at the centre of a circular coil of \( N \) turns is
\[ B = \frac{\mu_0 N I}{2r} \]

Step 3: Substitute the values.
\[ B = \frac{(4\pi\times10^{-7})(100)(1.0)}{2\times(2.0\times10^{-2})} \]

Step 4: Simplify.
Numerator \( = 4\pi\times10^{-7}\times100 = 4\pi\times10^{-5}\ \text{T m} \).
Denominator \( = 2\times2.0\times10^{-2} = 4.0\times10^{-2}\ \text{m} \).
\[ B = \frac{4\pi\times10^{-5}}{4.0\times10^{-2}} = \pi\times10^{-3}\ \text{T} \]

Step 5: Final value.
\[\boxed{B = \pi\times10^{-3}\ \text{T} \approx 3.14\times10^{-3}\ \text{T} = 3.14\ \text{mT}}\]
The field is directed along the axis of the coil (given by the right-hand rule).