A square shaped coil of area $$70$$ cm$$^2$$ having $$600$$ turns rotates in a magnetic field of $$0.4$$ Wb m$$^{-2}$$, about an axis which is parallel to one of the side of the coil and perpendicular to the direction of field. If the coil completes $$500$$ revolution in a minute, the instantaneous emf when the plane of the coil is inclined at $$60°$$ with the field, will be ______ V. (Take $$\pi = \frac{22}{7}$$)

A square-shaped coil of area 70 cm$$^2$$ with 600 turns rotates in a magnetic field of 0.4 Wb/m$$^2$$. The coil completes 500 revolutions per minute. Find the instantaneous emf when the plane of the coil is inclined at 60° with the field.

The coil has $$N = 600$$ turns, area $$A = 70$$ cm$$^2 = 70 \times 10^{-4}$$ m$$^2$$, and it rotates at $$f = \frac{500}{60}$$ rev/s in a magnetic field of strength $$B = 0.4$$ Wb/m$$^2$$. Here $$\pi = \frac{22}{7}$$.

The angular velocity is given by $$\omega = 2\pi f = 2 \times \frac{22}{7} \times \frac{500}{60} = \frac{22000}{420} = \frac{1100}{21}$$ rad/s.

The peak emf is $$\varepsilon_0 = NBA\omega = 600 \times 0.4 \times 70 \times 10^{-4} \times \frac{1100}{21}$$
$$= 600 \times 0.4 \times 0.007 \times \frac{1100}{21}$$
$$= 1.68 \times \frac{1100}{21} = 1.68 \times 52.381 = 88$$ V.

The instantaneous emf is $$\varepsilon = \varepsilon_0 \sin(\omega t)$$. When the plane of the coil is inclined at 60° with the field, the angle between the area vector and the field is $$90° - 60° = 30°$$. Thus, $$\varepsilon = \varepsilon_0 \sin(30°) = 88 \times \frac{1}{2} = 44$$ V.

The answer is $$44$$ V.