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A current of $I = \sqrt{2} A$ flows in a circuit having the shape of an isosceles trapezium. The ratio of the bases of the trapezium is 2. The length of the smaller base of the trapezium is l = 100 mm, the distance h =50mm. The magnetic induction B at symmetric point O in the plane of the trapezium is $x μ T$ . Then $' x'$ is

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Detailed Solution

$B_{2} = B_{4} = 0 at o$

$B_{1} = \frac{μ_{0} i}{4 π d} (\sin α + \sin β)$

$= \frac{1 0^{- 7} \times \sqrt{2}}{100 \times 10^{- 3}} (\sin 45 ° + \sin 45 °)$

$= \sqrt{2} \times 10^{- 6} \times 2 \frac{1}{\sqrt{2}} = 4 \times 10^{- 6} T$

$B_{3} = \frac{μ_{0} i}{4 π d} (\sin α + \sin β)$

$= \frac{10^{- 7} \sqrt{2}}{50 \times 10^{- 3}} 2 \times \frac{1}{\sqrt{2}} = 4 \times 10^{- 6} T$

$B_{0} = B_{3} - B_{1}$

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

$= 2 μ T$

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