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Q.

A circular arc of wire subtends an angle π/2 at the centre. If it carries a current i and its radius of curvature is R then the magnetic field at the centre of the arc is

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a

$\frac{μ_{0} i}{8 R}$

b

$\frac{μ_{0} i}{R}$

c

$\frac{μ_{0} i}{2 R}$

d

$\frac{μ_{0} i}{4 R}$

answer is D.

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Detailed Solution

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$B = \frac{μ_{0} i (θ)}{4 πr} where θ = \frac{π}{2}; B = \frac{μ_{0} iπ}{2 \times 4 πR} = \frac{μ_{0} i}{8 R}$

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