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Q.

A long wire having a semicircular loop of radius r carries a current i as shown in figure. The magnetic induction at the centre O due to entire wire is

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a

$\frac{μ_{0} {\dot{1}}^{2}}{4 r}$

b

$\frac{μ_{0} i}{2 r}$

c

$\frac{μ_{0} i}{4 r^{2}}$

d

$\frac{μ_{0} i}{4 r}$

answer is A.

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Detailed Solution

If the wire is divided in to three parts,

$B_{1} = 0$

$B_{2} = \frac{μ_{0} I}{4 π r} θ = \frac{μ_{0} I}{4 π r} π = \frac{μ_{0} I}{4 r}$

$B_{3} = 0$

$B = B_{1} + B_{2} + B_{3}$

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