Magnetic Force on Shaped Current Carrying Conductor

To determine the magnetic force acting on the conductor, we use the expression for the force on a current-carrying wire segment in a uniform magnetic field $\vec{B}$:

$\vec{F} = \int I (d\vec{l} \times \vec{B})$

Since the magnetic field $\vec{B}$ is uniform throughout the region, it can be taken outside the integral:

$\vec{F} = I \left( \int d\vec{l} \right) \times \vec{B} = I (\vec{L}_{\text{eff}} \times \vec{B})$

where $\vec{L}_{\text{eff}}$ is the effective displacement vector from the initial point $A$ to the final point $B$ of the conductor:

$\vec{L}_{\text{eff}} = \vec{r}_B - \vec{r}_A$

From the given figure:

• The conductor lies entirely in the $x-y$ plane.
• The initial point $A$ and final point $B$ both lie on the $x$-axis.
• The total horizontal displacement along the $+x$-direction is given by: $\Delta x = L + R + R + L = 2(L + R)$
• Since both endpoints lie on the $x$-axis, $\Delta y = 0$ and $\Delta z = 0$.

Thus, the effective displacement vector is:

$\vec{L}_{\text{eff}} = 2(L + R)\hat{i}$

Now, we evaluate the net force $\vec{F}$ and its magnitude $F$ for the different orientations of the uniform magnetic field $\vec{B}$:

1. If $\vec{B}$ is along $\hat{z}$ ($\vec{B} = B\hat{k}$): $\vec{F} = I \left[ 2(L + R)\hat{i} \times B\hat{k} \right] = -2IB(L + R)\hat{j}$ The magnitude of the force is: $F = 2IB(L + R)$ Therefore, $F \propto (L + R)$, which makes statement (A) correct and statement (D) incorrect.

2. If $\vec{B}$ is along $\hat{x}$ ($\vec{B} = B\hat{i}$): $\vec{F} = I \left[ 2(L + R)\hat{i} \times B\hat{i} \right] = \vec{0}$ The magnitude of the force is: $F = 0$ Therefore, statement (B) is correct.

3. If $\vec{B}$ is along $\hat{y}$ ($\vec{B} = B\hat{j}$): $\vec{F} = I \left[ 2(L + R)\hat{i} \times B\hat{j} \right] = 2IB(L + R)\hat{k}$ The magnitude of the force is: $F = 2IB(L + R)$ Therefore, $F \propto (L + R)$, which makes statement (C) correct.

Thus, the correct options are A, B, and C.