A current $I = \sqrt{2} A$ flows in a circuit having the shape of isosceles trapezium. The ratio of the bases of the trapezium is 2. Find the magnetic induction $B$ at symmetric point $O$ in the plane of the trapezium. The length of the smaller base of the trapezium is 100 mm and the distance r=50mm.
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$3 \times 10^{- 6} T$

$2 \times 10^{- 5} T$

$3 \times 10^{- 7} T$

$2 \times 10^{- 6} T$

answer is D.

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Detailed Solution

Magnetic fields at $O$ due to currents in wires $a b$ and $c d$ are zero.

Magnetic fields due to current in wire $d a$ $(s a y B_{2})$ is inwards due to current in wire $b c (s a y B_{1})$ is outwards.

$B_{1} = \frac{μ_{0}}{4 π} \frac{i}{r_{1}} (\sin α + \sin β) B_{2} = \frac{μ_{0}}{4 π} \frac{i}{r_{2}} (\sin α + \sin β) B_{1} > B_{2} a s r_{1} < r_{2} ∴ B_{n e t} = B_{1} - B_{2} (o u t w a r d s)$

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