NEET Physics Question (2026) — Solution

Step-by-step solution

From the given figure, the dashed line passing through A, O, and C is straight, indicating that the segments ABC and ADC are semicircles. Let the resistance of segment ADC be R_ ADC = 2R. Given that the resistance of segment ABC is half that of ADC, we have R_ ABC = R. The total current I_0 splits at point A into two parallel branches. Let I_1 be the current through ABC and I_2 be the current through ADC. Since the branches are in parallel, the potential difference across them is equal: I_1 R_ ABC = I_2 R_ ADC I_1 (R) = I_2 (2R) I_1 = 2I_2 Applying Kirchhoff's current law at junction A: I_1 + I_2 = I_0 2I_2 + I_2 = I_0 3I_2 = I_0 I_2 = I_0 3 Consequently, I_1 = 2I_0 3 . The magnetic field at the center of a semicircle carrying current I is given by B = _0 I 4r . For the left semicircle ABC, the current I_1 flows in a counter-clockwise direction. By the right-hand rule, its magnetic field at O is directed out of the plane: B_1 = _0 I_1 4r = _0 4r ( 2I_0 3 ) = _0 I_0 6r For the right semicircle ADC, the current I_2 flows in a clockwise direction. Its magnetic field at O is directed into the plane: B_2 = _0 I_2 4r = _0 4r ( I_0 3 ) = _0 I_0 12r The straight wire segments carrying current I_0 into A and out of C lie on the line passing through O, so they do not contribute to the magnetic field at O. The net magnetic field at the center O is the difference between B_1 and B_2 since they are in opposite directions: B_ net = B_1 - B_2 = _0 I_0 6r - _0 I_0 12r = 2 _0 I_0 - _0 I_0 12r = _0 I_0 12r Answer: _0 I_0 12r