A square shaped coil of area 70 cm² having 600 turns rotates in a magnetic field of 0.4 wbm⁻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)

Solution

Physics sir · 4 steps

Convert the plane's angle to the angle of the normal first, then e = NBAω sin θ does the rest.

1. Find the right angle. The flux is Φ = NBA cos θ, where θ is the angle between the field and the normal to the coil. The plane makes 60° with the field, so the normal makes θ = 90° - 60° = 30°.

   e = -dΦ/dt = NBAω sin θ

2. Put in numbers. A = 70 cm² = 7 × 10⁻³ m², so NBA = 600 × 0.4 × 7 × 10⁻³ = 1.68 Wb. Also ω = 2π × 500/60 = 2 × 22/7 × 25/3 = 1100/21 rad s⁻¹.
3. Multiply out. NBAω = 1.68 × 1100/21 = 88 V. Then e = 88 × sin 30° = 88 × 1/2.
4. So. The instantaneous emf is 44 V.

The common slip: 76.2. Putting the plane's angle straight into the formula, 88 × sin 60°, gives about 76.2 V instead of 44 V.

The idea to keep. In a rotating coil, θ in e = NBAω sin θ is measured from the normal, not the plane. Plane parallel to the field means maximum emf.

Answer from the NTA final answer key. Solution by PrepWiser.