A circular coil carrying current has 100 turns and a radius of 5.0 cm . It generates a magnetic field of 6.0 × 10^ -5 T at its centre. Determine the magnitude of the current.

Question

Quantrex Academy · Accepted Answer

Correct answer: B. 0.048 A

The magnetic field at the centre of a circular coil of

N

turns is given by:

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

Rearranging for current

I

:

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

Substituting the given values,

B = 6.0 \times 10^{-5} \text{ T}

,

R = 5.0 \text{ cm} = 0.05 \text{ m}

,

N = 100

, and

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

:

I = \dfrac{6.0 \times 10^{-5} \times 2 \times 0.05}{4\pi \times 10^{-7} \times 100}

I = \dfrac{6.0 \times 10^{-6}}{4\pi \times 10^{-5}}

I = \dfrac{0.15}{\pi} \text{ A}

Using

\pi \approx 3.14

:

I \approx \dfrac{0.15}{3.14} \approx 0.048 \text{ A}

A circular coil carrying current has 100 turns and a radius of 5.0...

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