Two long straight parallel wires are $2 m$ apart, perpendicular to the plane of the paper. The wire $A$ carries a current of $9.6 A$ , directed into the plane of the paper. The wire $B$ carries a current such that the magnetic field of induction at the point $P$ , at a distance of $\frac{10}{11} m$ form the wire $B$ , is zero. Find the magnitude of the magnetic field of induction at the point $S$ .
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$∠ A S B = 90 °$

$∠ A S B = 30 °$

$∠ A S B = 60 °$

$∠ A S B = 45 °$

answer is A.

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Detailed Solution

Since, $A S^{2} + B S^{2} = A B^{2} ∠ A S B = 90 °$

At $S : B_{1} =$ magnetic filed due to $i_{A}$

$= \frac{μ_{0}}{2 π} \frac{i_{A}}{1.6} = \frac{(2 \times 10^{- 7}) (9.6)}{1.6} = 12 \times 10^{- 7} T$

$B_{2}$ = magnetic field due to $i_{B}$

$= \frac{μ_{0}}{2 π} \frac{i_{B}}{1.2} = \frac{(2 \times 10^{- 7}) (3)}{1.2} = 5 \times 10^{- 7} T$

Small $B_{1}$ and $B_{2}$ are mutually perpendiuclar. Net magnetic field at $S$ would be

$B = \sqrt{B_{1} + B_{2}} = \sqrt{{(12 \times 10^{- 7})}^{2} + {(5 \times 10^{- 7})}^{2}} = 13 \times 10^{- 7} T$

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