The figure shows a circular loop made of a wire of radius R. The resistivity of the material varies as a function of  θ such that  ρ=ρ0sin2⁡θ. The positions of the Jockey such that the magnetic field at the centre (O) due to the current in the loop is zero, will be

Consider two sectors one of  α and other of (2π−α)B at O will be zero if I1α=I2(2π−α) ⇒I1=ER1 ,⇒ I2=ER2 So αR1=2π−αR2 ⇒R1=∫0αρ0sin2⁡θA Rd θ =ρ0RAα-sin2α2∵∫sin2θdθ=12x-sinθcosθ+CSimilarly  we can get  R2=∫α2πρ0sin2⁡θA Rd θ =ρ0RA2π-α+sin2α2 ∴αR2=2π-αR1⇒αρ0RA2π-α+sin2α2=2π-αρ0RAα-sin2α2 ⇒sin2α=0Onsolvingweget  α=π2,π,3π2