For any arbitrary curved conductor carrying steady current $I$ placed in a uniform magnetic field $\vec{B}$, the net magnetic force depends solely on the vector displacement $\vec{L}$ connecting the initial and final endpoints:

• The wire starts at point P and ends at point Q. With two semicircles each of radius $R$ spanning horizontally, the net displacement vector from P to Q along the horizontal $x$-axis is:
  $$\vec{L} = (2R + 2R)\,\hat{i} = 4R\,\hat{i}$$
• The uniform magnetic field emerges normally out of the plane along the $+z$-axis:
  $$\vec{B} = B\,\hat{k}$$
• Taking the cross product:
  $$\vec{F} = I (4R\,\hat{i} \times B\,\hat{k}) = 4 B I R (\hat{i} \times \hat{k}) = 4 B I R (-\hat{j}) = -4 B I R\,\hat{j}$$
  This corresponds to a magnitude of $4BIR$ directed vertically downward.

• For a straight thin bar magnet of length $L$ and pole strength $m$, the initial magnetic dipole moment is:
  $$M = m L$$
• When the magnet is bent into a semicircle of radius $r$, the arc length equals the initial length:
  $$\pi r = L \implies r = \frac{L}{\pi}$$
• The new magnetic length is the straight-line distance between the two poles (the diameter of the semicircle):
  $$L' = 2r = 2\left(\frac{L}{\pi}\right) = \frac{2L}{\pi}$$
• Since the pole strength $m$ remains unchanged:
  $$M' = m L' = m \left(\frac{2L}{\pi}\right) = \frac{2(mL)}{\pi} = \frac{2M}{\pi}$$

Transformer cores undergo repeated cycles of magnetization and demagnetization at the supply frequency (e.g., 50 Hz):

• High magnetic permeability ($\mu_r$) is required so that large magnetic flux density is induced for a small magnetizing current, ensuring efficient flux linkage between primary and secondary windings.
• Low hysteresis loss (narrow hysteresis loop area) and low coercivity are required to minimize heat generation and energy dissipation during continuous cyclic magnetization.
  Soft iron or silicon steel is selected for transformer cores because it satisfies both criteria.

Hence, option 2 is the correct answer.