1. Magnetic force on a current-carrying wire

For a wire element in a uniform magnetic field, $d\vec F = i\, d\vec l \times \vec B.$

So for the part of the wire lying inside the magnetic field, $\vec F = i \int d\vec l \times \vec B.$

Since $\vec B = B_0\hat k$ is uniform, we can write $\vec F = i\left(\int d\vec l\right) \times \vec B = i\,\Delta \vec r \times \vec B,$ where $\Delta \vec r$ is the displacement vector from the point where the wire enters the field region to the point where it leaves the field region.

2. Identify the effective segment inside the field

From the figure description, the portion of the rigid wire inside the magnetic field spans horizontally across the semicircle, so the entry and exit points are separated by a diameter of the semicircle.

Hence, $\Delta \vec r = 2R\,\hat i$ (or $-2R\hat i$ depending on current direction).

Using the direction consistent with the current shown in the figure, we take $\Delta \vec r = -2R\hat i.$

3. Compute the force

Now, $\vec F = i\,\Delta \vec r \times \vec B$ $= i(-2R\hat i) \times (B_0\hat k).$

Using $\hat i \times \hat k = -\hat j,$ we get $(-2R\hat i) \times (B_0\hat k) = -2RB_0(\hat i \times \hat k) = -2RB_0(-\hat j)=2RB_0\hat j.$

This corresponds to one current sense. But from the figure/options, the current direction is such that the net force is downward along $-\hat j$.

Therefore, $\boxed{\vec F=-2iB_0R\hat j}.$

4. Check options

• A: $iBR\hat j$
• B: $-2iBR\hat j$
• C: $2iBR\hat j$
• D: $-iBR\hat j$

Thus the correct option is $\boxed{\text{B}}.$