Side of the square loop $a = 50\text{ cm} = 0.5\text{ m}$. Area $A = a^2 = (0.5)^2 = 0.25\text{ m}^2$. Magnetic field $B = 3.0\text{ T}$. Time interval $\Delta t = 0.3\text{ s}$.
Initial flux $\phi_i = B \cdot A \cdot \cos(0^{\circ}) = 3.0 \times 0.25 \times 1 = 0.75\text{ Wb}$. Since the loop is rotated by $90^{\circ}$, the final flux $\phi_f = B \cdot A \cdot \cos(90^{\circ}) = 0\text{ Wb}$.
The induced emf is given by the magnitude of the rate of change of magnetic flux: $$|\varepsilon| = \left| \frac{\Delta \phi}{\Delta t} \right| = \left| \frac{\phi_f - \phi_i}{\Delta t} \right|$$
Substituting the values: $$|\varepsilon| = \left| \frac{0 - 0.75}{0.3} \right| = \frac{0.75}{0.3} = 2.5\text{ V}$$
Final Answer: 2.5 V
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