A square loop of side \(2\) cm is placed in a time varying magnetic field with magnitude as \(B = 0.4\sin(300t)\) Tesla. The normal to the plane of loop makes an angle of \(60°\) with the field. The maximum induced emf produced in the loop is __________ mV.

Given:
Side of the square loop, \(a = 2\) cm \(= 2 \times 10^{-2}\) m
Area of the loop, \(A = a^2 = 4 \times 10^{-4}\) m\(^2\)
Magnetic field, \(B = 0.4 \sin(300t)\) T
Angle between the normal to the plane and the magnetic field, \(\theta = 60^\circ\)

The magnetic flux through the loop is given by:
\(\Phi = \vec{B} \cdot \vec{A} = B A \cos\theta\)
\(\Phi = (0.4 \sin(300t))(4 \times 10^{-4}) \cos(60^\circ)\)
\(\Phi = 1.6 \times 10^{-4} \sin(300t) \times \dfrac{1}{2}\)
\(\Phi = 0.8 \times 10^{-4} \sin(300t)\) Wb

By Faraday's law of induction, the induced emf is:
\(e = -\dfrac{d\Phi}{dt}\)
\(e = -\dfrac{d}{dt} (0.8 \times 10^{-4} \sin(300t))\)
\(e = -0.8 \times 10^{-4} \times 300 \cos(300t)\)
\(e = -240 \times 10^{-4} \cos(300t)\) V
\(e = -24 \times 10^{-3} \cos(300t)\) V

The maximum induced emf is the amplitude of \(e\):
\(e_{\text{max}} = 24 \times 10^{-3}\) V \(= 24\) mV

Answer: \(24\)