A conducting wire is bent into a circular arc of radius 2 cm and subtends an angle of 60^ at its centre. If a steady current of 12 A is passed through the wire, what is the magnitude of the magnetic field produced at

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Correct answer: B. 20π

The magnetic field at the centre of a circular arc is given by the formula:

B = \dfrac{\mu_0 I}{4\pi R} \theta

Given:

I = 12\text{ A}

R = 2\text{ cm} = 2 \times 10^{-2}\text{ m}

\theta = 60^\circ = \dfrac{\pi}{3}\text{ rad}

\mu_0 = 4\pi \times 10^{-7}\text{ T m A}^{-1}

Substituting the values into the formula:

B = \dfrac{4\pi \times 10^{-7} \times 12}{4\pi \times 2 \times 10^{-2}} \times \dfrac{\pi}{3}

B = \dfrac{10^{-7} \times 12}{2 \times 10^{-2}} \times \dfrac{\pi}{3}

B = 6 \times 10^{-5} \times \dfrac{\pi}{3}

B = 2\pi \times 10^{-5}\text{ T}

B = 20\pi \times 10^{-6}\text{ T} = 20\pi\text{ }\mu\text{T}

Answer:

20\pi\text{ }\mu\text{T}

A conducting wire is bent into a circular arc of radius 2 cm and...

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