JEE MainPhysicsModerateMagnetic effect of current

A conducting circular loop of copper is placed as shown in
figure. Cross section area of part abc is A and part adc is A/3 then
magnetic field at point is

Show answer and solution

R = \frac{\rho l}{A} \Rightarrow \frac{R_{abc}}{R_{adc}} = \left( \frac{\mathcal{l}}{A} \right)_{abc} \times \left( \frac{A}{\mathcal{l}} \right)_{adc}

= \frac{3/4\mathcal{l}}{A} \times \frac{A/3}{\mathcal{l/}4} = \frac{1}{1}

\frac{R_{abc}}{R_{adc}} = \frac{1}{1} \Rightarrow I_{abc} = I_{adc} = \frac{I}{2}

B_{0} = \frac{\mu_{0}I_{abc}\alpha_{1}}{4\pi R} - \frac{\mu_{0}I_{adc}\alpha_{2}}{4\pi R}

= \frac{\mu_{0}\left( \frac{I}{2} \right)}{4\pi R}\left( \alpha_{1} - \alpha_{2} \right) = \frac{\mu_{0}\left( \frac{I}{2} \right)}{4\pi R}\left( \frac{3\pi}{2} - \frac{\pi}{2} \right)

= \frac{\mu_{0}I}{8R} \otimes

Source: Grass-G / P-13 Physics paper 4 Nov.docx

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