Q.A current carrying circular loop of radius $R$ is placed in the $x$-$y$ plane with centre at the origin. Half of the loop with $x > 0$ is now bent so that it now lies in the $y$-$z$ plane.

Bending the $x>0$ half into the $y$-$z$ plane turns the loop into two perpendicular semicircular arcs, each of moment $m_0=I\pi R^2/2$ - one along $\hat{k}$, one along $\hat{i}$.

1. Net moment: $|\vec{M}|=m_0\sqrt{2}=\dfrac{I\pi R^2}{\sqrt{2}}\approx0.71\,I\pi R^2$ - smaller than the original $I\pi R^2$, so the magnitude diminishes (option a true, b false).
2. Far-axial field at $(0,0,z)$, using the dipole formula with $\vec{M}\cdot\hat{z}=m_0$, gives $|\vec{B}|=\dfrac{\mu_0 m_0}{4\pi z^3}\sqrt{5}$, while the original field was $\dfrac{\mu_0 m_0}{\pi z^3}$ - a ratio of $\sqrt{5}/4\approx0.56$, so the field also decreases (options c, d both false).

Before bending

A full circular loop of radius $R$ carrying current $I$ has magnetic moment

After bending

The loop now consists of two semicircular arcs of radius $R$, joined along the diameter on the $y$-axis:

• the $x<0$ half stays in the $x$-$y$ plane, contributing a (half-loop) magnetic moment $m_0=\dfrac{I\pi R^2}{2}$ along $\hat{k}$;
• the $x>0$ half is bent into the $y$-$z$ plane, contributing $m_0=\dfrac{I\pi R^2}{2}$ along $\hat{i}$ (its own normal direction, once it lies in the $y$-$z$ plane).

(Each semicircular arc encloses half the area of the full circle, $\pi R^2/2$, so its moment is half of $M_0$.)

Net magnetic moment - the two contributions are perpendicular, so they add as vectors:

Since $0.71\,M_0 < M_0$, the magnitude of the magnetic moment diminishes - option (a) is true, and option (b) ("does not change") is false.

Far-axial field at $(0,0,z)$, $z\gg R$

For a point far from a magnetic dipole $\vec{M}$, at position vector $\vec{r}$ ($\hat{r}=\hat{z}$ here, $r=z$):

With $\vec{M}=m_0(\hat{i}+\hat{k})$, $\vec{M}\cdot\hat{z}=m_0$:

Compare with the original far-axial field (dipole $M_0=2m_0$ along $\hat{k}$, on-axis):

The ratio is

so the field decreases - options (c) ("increases") and (d) ("unchanged") are both false.

Method: Vector Superposition of Magnetic Dipole Moments for Bent/Split Loops

When a loop is cut, bent out of its original plane, or split into pieces carrying the same current, treat each piece as its own magnetic dipole and add the dipole moment vectors -- never just rescale the original scalar moment.

Steps

Step 1: Find the dipole moment of each piece separately

Each planar piece of wire, carrying current $I$ and enclosing area $A_i$, has its own magnetic moment of magnitude $IA_i$, directed along that piece's own normal (right-hand rule applied to that piece alone):

A semicircular half of a full loop of radius $R$ encloses half the area, so $|\vec m_i| = I(\pi R^2/2)$.

Step 2: Add the pieces as vectors, not as scalars

If the pieces now lie in different planes (as after bending part of a loop out of its original plane), their normals $\hat n_i$ point in different directions, so the net moment is a genuine vector sum:

This magnitude is generally less than the simple scalar sum $\sum_i|\vec m_i|$ whenever the pieces are not all parallel -- this is the source of "the moment diminishes" answers in this class of problem.

Step 3: Use the far-axial dipole field with the new $\vec M$

Once $\vec M$ is known, the field far from the dipole along a given axis (here the $z$-axis) follows the standard axial dipole formula:

Compare $|\vec B|$ before and after bending by taking the ratio -- the geometry (perpendicular vs. parallel components of $\vec M$) usually matters more than the raw magnitude of $\vec M$.

Applying to this problem: the two perpendicular semicircular moments ($m_0\hat k$ and $m_0\hat i$) combine to $|\vec M|=m_0\sqrt2\approx0.71\,I\pi R^2$ (Step 2), smaller than the original $I\pi R^2$, and the resulting far-axial field is correspondingly smaller too (Step 3).