Find the magnetic field due to conducting wire at point $O,$ at centre of semicircle of radius $r$ and carrying a current $i$ as shown in the figure.

\frac{μ_{0} i}{4 r}

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\frac{μ_{0} i}{4 r} (1 + 2 π)

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\frac{μ_{0} i}{4 r} (π - 2)

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\frac{μ_{0} i}{4 π r} (π + 2)

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Solution

The correct option is B $\frac{μ_{0} i}{4 π r} (π + 2)$
$| B |$ at centre = $| B |$ due to semicircle wire+ $2 \times$ $| B |$ due semi infinite wire
$= \frac{μ_{0} i}{4 r} + \frac{2 μ_{0} i}{4 π r}$
$= \frac{μ_{0} i}{4 r} \times (1 + \frac{2}{π})$

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Find the magnetic field due to conducting wire at point O, at centre of semicircle of radius r and carrying a current i as shown in the figure.

The correct option is B μ0i4πr(π+2)|B| at centre =|B| due to semicircle wire+ 2× |B| due semi infinite wire=μ0i4r+2μ0i4πr=μ0i4r×(1+2π)

Find the magnetic field due to conducting wire at point $O,$ at centre of semicircle of radius $r$ and carrying a current $i$ as shown in the figure.

The correct option is B $\frac{μ_{0} i}{4 π r} (π + 2)$
$| B |$ at centre = $| B |$ due to semicircle wire+ $2 \times$ $| B |$ due semi infinite wire
$= \frac{μ_{0} i}{4 r} + \frac{2 μ_{0} i}{4 π r}$
$= \frac{μ_{0} i}{4 r} \times (1 + \frac{2}{π})$