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A long wire i...

Question

A long wire is bent into the shape PQRST as shown in the following figure with QRS being a semicircle with Centre O and radius r meter.
A current of I A flows through it in the direction as shown in the figure. Then the magnetic induction at the point O of the figure in vacuum is:

\frac{μ_{0} I}{2 r} [\frac{1}{π} + \frac{1}{2}]

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\frac{μ_{0} I}{2 r} [\frac{1}{π} - \frac{1}{2}]

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\frac{μ_{0} I}{4 r}

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\frac{μ_{0} I}{π r}

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Solution

The correct option is A $\frac{μ_{0} I}{2 r} [\frac{1}{π} + \frac{1}{2}]$
$B_{1} = \frac{μ_{0} i}{4 π R} (-^k); B_{2} = \frac{μ_{0} i}{4 R} (-^k)$
$B_{3} = \frac{μ_{0} i}{4 π R} (-^k); B_{n e t} = B_{1} + B_{2} + B_{3}$ .

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A long wire is bent into the shape PQRST as shown in the following figure with QRS being a semicircle with Centre O and radius r meter. A current of I A flows through it in the direction as shown in the figure. Then the magnetic induction at the point O of the figure in vacuum is:

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A long wire is bent into the shape PQRST as shown in the following figure with QRS being a semicircle with Centre O and radius r meter.
A current of I A flows through it in the direction as shown in the figure. Then the magnetic induction at the point O of the figure in vacuum is:

The correct option is A $\frac{μ_{0} I}{2 r} [\frac{1}{π} + \frac{1}{2}]$
$B_{1} = \frac{μ_{0} i}{4 π R} (-^k); B_{2} = \frac{μ_{0} i}{4 R} (-^k)$
$B_{3} = \frac{μ_{0} i}{4 π R} (-^k); B_{n e t} = B_{1} + B_{2} + B_{3}$ .