A rectangular loop of length $$2.5$$ m and width $$2$$ m is placed at $$60°$$ to a magnetic field of $$4$$ T. The loop is removed from the field in $$10$$ sec. The average emf induced in the loop during this time is

A rectangular loop of length 2.5 m and width 2 m is placed at an angle of 60° with a magnetic field of 4 T and is removed from the field in 10 s.

The magnetic flux through the loop initially is given by $$\Phi = BA\cos\theta = 4 \times (2.5 \times 2) \times \cos 60° = 4 \times 5 \times 0.5 = 10$$ Wb.

According to Faraday’s law, the average induced emf is $$\mathcal{E} = -\frac{\Delta\Phi}{\Delta t} = -\frac{0 - 10}{10} = \frac{10}{10} = +1$$ V.

The correct answer is Option C: +1 V.