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A current carrying wire carrying current is bent in the form of a circular arc which subtends an angle of $60^{o}$ at its centre. If length of the wire is l, magnetic induction at the centre of curvature of the bent wire is

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$30 π i / 9 l$

$μ_{0} π i / 36 l$

$\frac{μ_{0} i}{36 π l}$

$\frac{μ_{0} i}{18 π l}$

answer is D.

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Detailed Solution

$R = \frac{l}{θ}$

$∴ B_{0} = \frac{μ_{0} i}{2 r} . \frac{θ}{2 π}$

$= \frac{μ_{0} i}{(2 l / θ)} \frac{θ}{2 π} = \frac{μ_{0} i}{4 π l} . {(\frac{π}{3})}^{2} = \frac{μ_{0} π i}{36 l}$

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