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प्रश्न
A straight wire carrying a current of 5 A is bent into a semicircular arc of radius 2 cm as shown in the figure. Find the magnitude and direction of the magnetic field at the centre of the arc.
थोडक्यात उत्तर
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उत्तर
Current I = 5A, radius r = 2 cm = 2 x 10-2 m, Number of turns = N = 1/2
So, magnetic field due to semi-circular arc is
`b = (mu_0 NI)/(2r) = (4 pi xx 10^(-7) xx 0.5 xx5)/(2 xx 2 xx 10^(-2)) T = 7.85 xx 10^(-5) T `
The magnetic field due to the straight part of the wire is zero at the centre as its line intersects the centre.
Hence net magnetic field = 7.85×10-5T
The direction of the this magnetic field is perpendicular to the plane of the paper and directed inside .
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