A circular coil consisting of 200 turns has a radius of 10 cm and carries a current of 2.0 A . Determine the magnitude of the magnetic field B at the centre of the coil.

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Quantrex Academy · Accepted Answer

Correct answer: A. 2.51 mT

The magnetic field at the centre of a circular coil of

N

turns is given by:

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

Given parameters:

N = 200

R = 10 \text{ cm} = 0.1 \text{ m}

I = 2.0 \text{ A}

\mu_0 = 4\pi \times 10^{-7} \text{ T}\cdot\text{m/A}

Substituting the values into the formula:

B = \dfrac{4\pi \times 10^{-7} \times 200 \times 2.0}{2 \times 0.1}

B = \dfrac{1600\pi \times 10^{-7}}{0.2}

B = 8000\pi \times 10^{-7} \text{ T}

B = 8\pi \times 10^{-4} \text{ T}

Using

\pi \approx 3.14

:

B = 8 \times 3.14 \times 10^{-4} \text{ T}

B = 25.12 \times 10^{-4} \text{ T}

B = 2.51 \times 10^{-3} \text{ T} = 2.51 \text{ mT}

A circular coil consisting of 200 turns has a radius of 10 cm and...

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