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A piece of wire carrying a current of 6A is bent in the form of a circular arc of radius 10cm and it subtends an angle of 1200 at the centre. Find the magnetic field B due to this piece of wire at the centre.

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$0 .5 \times 10^{- 5} T$

$1 .26 \times 10^{- 5} T$

$2 .4 \times 10^{- 5} T$

$3 .2 \times 10^{- 5} T$

answer is B.

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Detailed Solution

$\begin{array}{l} B_{c} = \frac{μ_{0} i}{2 a} (\frac{θ}{2 π}), θ = 120^{0} = 2 π / 3 \\ B_{c} = \frac{4 π \times 10^{7} \times 6}{2 \times (0 .1)} (\frac{2 π}{3 \times 2 π}) \\ = 1 .256 \times 10^{5} T \end{array}$

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