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Five long wires $A,B,C,D$ and $E$ each carrying current I are arranged to form edges of a pentagonal prism as shown in figure. Each carries current out of the plane of paper . What will be the field if current in one of the wire (say A) is switched off ?

$\begin{array}{1 1} \large\frac{\mu_0}{4 \pi} .\frac{2I}{R} \text{ acting perpendicular to OA towards right} \\ \large\frac{\mu_0}{4 \pi} .\frac{2I}{R} \text{ acting perpendicular to AO towards right} \\ \large\frac{\mu_0}{4 \pi} .\frac{2I}{R} \text{ acting perpendicular to OA towards left} \\ \large\frac{\mu_0}{4 \pi} .\frac{2I}{R} \text{ acting perpendicular to AO towards left} \end{array} $

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A)
Solution :
Total magnetic field induction at 0 due to current through wires B,C,D and E are equal and opposite due to current through wire A.
Magnetic field induction at 0 due to wire A is
$B= \large\frac{\mu_0}{4 \pi }. \frac{2I}{R}$
Which acts perpendicular to AO towards right .
Therefore magnetic field induction through the wires $B,C,D,E$ is $\large\frac{\mu_0}{4 \pi} .\frac{2I}{R}$ acting perpendicular to AO towards left
Answer : $\large\frac{\mu_0}{4 \pi} .\frac{2I}{R}$ acting perpendicular to AO towards left
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