A wire carrying current I, bent as shown in the figure, is placed in a uniform mag netic field B that emerges normally out from the plane of the figure. The force on this wire is ________.

Fill in the blank with the correct answer from the options given below.

The correct option is 4BIR, directed vertically downward.

1. Understanding the Magnetic Force on a Current-Carrying Wire:
When a wire of any arbitrary shape carrying a current $I$ is placed in a uniform magnetic field $\overrightarrow{B}$, the net magnetic force acting on the wire is equivalent to the force acting on a straight wire connecting its endpoints. The formula is:

$\overrightarrow{F}=I({\overrightarrow{L}}_{\mathrm{eff}}\times \overrightarrow{B})$

where:
• $I$ is the electric current in the wire,
• ${\overrightarrow{L}}_{\mathrm{eff}}$ is the effective length vector, which is the displacement vector pointing directly from the starting point P to the ending point Q, and
• $\overrightarrow{B}$ is the uniform magnetic field vector.

2. Analyzing the Geometry of the Wire:
Looking at the provided diagram, the wire is bent into three distinct sections:
• A left horizontal straight segment of length $R$.
• A central semicircular arc of radius $R$. The horizontal distance between the start and end of this semicircle is its diameter, which is equal to $2R$.
• A right horizontal straight segment of length $R$.

3. Calculating the Effective Length Vector:
The net displacement vector starts at P and ends at Q along a straight horizontal path. The magnitude of this effective length vector is the sum of the horizontal spans of the three parts:

$L_{\mathrm{eff}}=R+2R+R=4R$

If we define the horizontal direction from left to right as the positive x-axis ($\hat{i}$), the effective length vector is:

${\overrightarrow{L}}_{\mathrm{eff}}=4R\hat{i}$

4. Computing the Magnetic Force:
The uniform magnetic field $\overrightarrow{B}$ is directed normally out of the plane of the figure, which represents the positive z-axis ($\hat{k}$):

$\overrightarrow{B}=B\hat{k}$

Substituting these vectors into the force equation:

$\overrightarrow{F}=I(4R\hat{i}\times B\hat{k})$

$\overrightarrow{F}=4BIR(\hat{i}\times \hat{k})$

Using the cross product rule for unit vectors, we have $\hat{i}\times \hat{k}=-\hat{j}$ (where $-\hat{j}$ represents the direction pointing vertically downward):

$\overrightarrow{F}=-4BIR\hat{j}$

Thus, the magnitude of the force is 4BIR, and its direction is vertically downward.