A square shaped coil of area $$70 \text{ cm}^2$$ having 600 turns rotates in a magnetic field of $$0.4 \text{ Wb m}^{-2}$$, 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 $$\pi = \frac{22}{7}$$)

We need to find the instantaneous emf of a rotating coil when the plane is inclined at 60° with the field.

The coil has area $$A = 70$$ cm², which is $$70 \times 10^{-4}$$ m²; it consists of $$N = 600$$ turns, rotates at 500 rpm in a magnetic field of $$B = 0.4$$ T, and we take $$\pi = 22/7$$.

Since the coil rotates at 500 rpm, its angular frequency is given by $$\omega = \frac{2\pi \times 500}{60} = \frac{1000 \times 22}{60 \times 7} = \frac{22000}{420} = \frac{1100}{21}$$ rad/s.

When the plane of the coil is inclined at 60° to the field, the normal to the coil makes an angle of $$90° - 60° = 30°$$ with the field. The induced emf is given by $$\varepsilon = NBA\omega\sin(\omega t)$$ where $$\omega t = 30°$$.

Substituting the values into the expression for $$NBA\omega$$ gives $$NBA\omega = 600 \times 0.4 \times 70 \times 10^{-4} \times \frac{1100}{21} = 600 \times 0.4 \times 0.007 \times \frac{1100}{21} = 1.68 \times \frac{1100}{21} = \frac{1848}{21} = 88$$ V.

Then $$\varepsilon = 88 \times \sin 30° = 88 \times 0.5 = 44$$ V.

Therefore, the answer is 44.