The user wants me to solve a physics problem about the magnetic force on a current-carrying conductor.

Step-by-step Solution

1. Identify the governing principle. The total magnetic force $\vec{F}$ on a conductor carrying a constant current I from a starting point A to an ending point B in a uniform magnetic field $\vec{B}$ can be calculated using the effective length vector $\vec{L}_{eff}$. The formula is: $\vec{F} = I (\vec{L}_{eff} \times \vec{B})$ where $\vec{L}_{eff}$ is the vector pointing from the start point A to the end point B of the conductor.

2. Determine the effective length vector $\vec{L}_{eff}$. From the provided figure, the conductor starts at the origin, A = (0, 0, 0). The current flows through a path consisting of a straight section, a semicircular section, and another straight section, finally ending at the point B = (2R, 0, 0). The effective length vector is the displacement vector from A to B: $\vec{L}_{eff} = \text{Position B} - \text{Position A} = (2R\hat{i} + 0\hat{j} + 0\hat{k}) - (0\hat{i} + 0\hat{j} + 0\hat{k}) = 2R\hat{i}$

3. Establish the general formula for the force on this conductor. Substituting $\vec{L}_{eff}$ into the force equation, we get: $\vec{F} = I (2R\hat{i} \times \vec{B})$

4. Evaluate each option based on this formula.

   • A: if B is along $\widehat z$ ($\vec{B} = B\hat{k}$), F $\propto$ (L + R) $\vec{F} = I (2R\hat{i} \times B\hat{k}) = 2IRB (\hat{i} \times \hat{k})$ Using the cross-product rule $\hat{i} \times \hat{k} = -\hat{j}$: $\vec{F} = -2IRB\hat{j}$ The magnitude of the force is $F = |\vec{F}| = 2IRB$. This shows that F is proportional to R ($F \propto R$). The statement $F \propto (L + R)$ is incorrect because L is an independent parameter.

   • B: if B is along $\widehat x$ ($\vec{B} = B\hat{i}$), F = 0 $\vec{F} = I (2R\hat{i} \times B\hat{i}) = 2IRB (\hat{i} \times \hat{i})$ The cross product of a vector with itself is the zero vector, $\hat{i} \times \hat{i} = 0$. $\vec{F} = 0$ The magnitude is F = 0. This statement is correct.

   • C: if B is along $\widehat y$ ($\vec{B} = B\hat{j}$), F $\propto$ (L + R) $\vec{F} = I (2R\hat{i} \times B\hat{j}) = 2IRB (\hat{i} \times \hat{j})$ Using the cross-product rule $\hat{i} \times \hat{j} = \hat{k}$: $\vec{F} = 2IRB\hat{k}$ The magnitude of the force is $F = |\vec{F}| = 2IRB$. This shows that F is proportional to R ($F \propto R$). The statement $F \propto (L + R)$ is incorrect.

   • D: if B is along $\widehat z$ ($\vec{B} = B\hat{k}$), F = 0 From the analysis for option A, the force magnitude is F = 2IRB. This is non-zero (assuming $I, R, B \neq 0$). This statement is incorrect.

Conclusion

Based on a rigorous physical analysis, only option B is correct. Options A and C are incorrect because the net force on the conductor is proportional to R, not (L+R). The forces on the two straight sections of length L either cancel each other out (in case A) or are zero (in case C), so L does not appear in the final expression for the net force.

This is a known problematic question from a past competitive exam (JEE Advanced 2015). The official answer key marks A, B, and C as correct. This is likely due to a typo in options A and C, where (L+R) should have been R. If the options were $F \propto R$, then both A and C would be correct, as the magnitude of the force is 2IRB in both cases. However, as the question is written, only option B is physically correct. My derived answer is based on the question as stated.