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 π = 22 7 )

Step-by-step solution

Given here, N = 600 , A = 70 × 10 - 4   m 2 , and B = 0 . 4   T Angular speed of coil is ω = 2 π n 60 = 500 × 2 π 60 = 100 π 6   rad   s - 1 Instantaneous emf in coil is given by ε = - N d ϕ d t , where, magnetic flux ϕ = B A cos ω t and ω t = 30 ° is angle between A and B . So, ε = - N d B A cos ω t d t = - N - B A ω sin ω t = N B A ω sin ω t Putting the values, we get ε = 600 × 70 × 10 - 4