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JEE Main 2020 · Physics · Magnetic Effects of Current · medium

A wire carrying a current I forms a closed loop made of two concentric quarter-circular arcs of radii R and 2R, centred at O and lying between the…

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A wire carrying a current II forms a closed loop made of two concentric quarter-circular arcs of radii RR and 2R2R, centred at OO and lying between the same two radii, joined by two straight radial segments. The current flows in one sense (say anticlockwise) along the outer arc and therefore in the opposite sense along the inner arc. The magnitude of the magnetic field at OO is
  1. A. 3μ0I16R\dfrac{3\mu_0 I}{16R}
  2. B. μ0I16R\dfrac{\mu_0 I}{16R}
  3. C. μ0I8R\dfrac{\mu_0 I}{8R}
  4. D. μ0I4R\dfrac{\mu_0 I}{4R}
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Answer: (B) μ0I16R\dfrac{\mu_0 I}{16R}

The radial segments pass through OO, so they contribute nothing. An arc subtending angle θ\theta gives B=μ0Iθ4πrB = \dfrac{\mu_0 I\theta}{4\pi r} at the centre; with θ=π/2\theta = \pi/2: inner arc B1=μ0I8RB_1 = \dfrac{\mu_0 I}{8R}, outer arc B2=μ0I16RB_2 = \dfrac{\mu_0 I}{16R}. The senses are opposite, so the fields oppose: B=μ0I8R−μ0I16R=μ0I16RB = \dfrac{\mu_0 I}{8R} - \dfrac{\mu_0 I}{16R} = \dfrac{\mu_0 I}{16R}. The sum 3μ0I/16R3\mu_0 I/16R would result if the senses were the same.
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