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A wire length $\frac{π}{2} m$ , carrying a current of 2A is bent in the form of a circular arc which subtends on angle of 45° at its centre. Then magnetic induction at the centre is

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µ_o / 8 T

µ_o / 16 T

µ_o / 2 T

4 µ_o T

answer is B.

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Detailed Solution

$l = R . θ \Rightarrow R = \frac{π / 2}{π / 4} m = 2 m ∴ B_{o} = \frac{μ_{o} i}{2 R} . \frac{θ}{2 π} = \frac{μ_{o} \times 2}{2 \times 2} \times \frac{π / 4}{2 π} T \Rightarrow B_{o} = \frac{μ_{o}}{16} T$

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