A uniform straight bar magnet has a magnetic dipole moment μ . It...

Let the initial length of the bar magnet be $2l$ and its pole strength be $m$.

The initial magnetic dipole moment is given by $\mu = m(2l)$.

When the magnet is bent at its exact center, it forms two arms of equal length $l$. The angle between these two arms is given as $120^{\circ}$.

The new distance $d$ between the North and South poles can be calculated using the cosine rule in the triangle formed by the two arms:

$d = \sqrt{l^2 + l^2 - 2(l)(l)\cos(120^{\circ})}$

Substituting $\cos(120^{\circ}) = -\dfrac{1}{2}$:

$d = \sqrt{2l^2 - 2l^2\left(-\dfrac{1}{2}\right)} = \sqrt{2l^2 + l^2} = \sqrt{3}l$

The new magnetic dipole moment $\mu'$ is the product of the pole strength and the new distance between the poles:

$\mu' = m(\sqrt{3}l) = \dfrac{\sqrt{3}}{2}(m \times 2l)$

Substituting the initial magnetic moment $\mu = 2ml$:

$\mu' = \dfrac{\sqrt{3}}{2}\mu$

Answer: $\dfrac{\sqrt{3}}{2}\mu$