A square coil of 10 cm side and with 100 turns is rotated at a uniform

1.	A square coil of 10 cm side and with 100 turns is rotated at a uniform speed of 500 rpm about an axis at right angle to a uniform field of 0.5 Wb/m2 . Calculate the instantaneous value of induced e.m.f. when the plane of the coil is (i) at right angle to the plane of the field. (ii) in the plane of the field. (iii) at 45º with the field direction.
Answer» As seen from , e.m.f. induced in the coil would be zero when its plane is at right angles to the plane of the field, even though it will have maximum flux linked with it. However, the coil will have maximum e.m.f. induced in it when its plane lies parallel to the plane of the field even though it will have minimum flux linked with it. In general, the value of the induced e.m.f. is given by e = ωNΦ_m sinθ = E_m sinθ where θ is the angle between the axis of zero e.m.f. and the plane of the coil. Here, f = 500/ 60 = 25/ 3 r.p.s ; N = 100 ; B = 0.5 Wb/ m² ; A = (10 × 10) × 10⁻⁴ = 10⁻² m² . ∴ E_m= 2π f NBA = 2π (25/3) × 100 × 0.5 × 10⁻² = 26.2 V (i) since θ = 0 ; sin θ = 0 ; therefore, e = 0. (ii) Here, θ = 90° ; e = E_m sin 90º = 26.2 × 1 = 26.2 V (iii) sin 45º = 1/√2 ; e = 26.2 × 1/√2 = 18.5 V

A square coil of 10 cm side and with 100 turns is rotated at a uniform speed of 500 rpm about an axis at right angle to a uniform field of 0.5 Wb/m2 . Calculate the instantaneous value of induced e.m.f. when the plane of the coil is (i) at right angle to the plane of the field. (ii) in the plane of the field. (iii) at 45º with the field direction.

Answer»

As seen from , e.m.f. induced in the coil would be zero when its plane is at right angles to the plane of the field, even though it will have maximum flux linked with it. However, the coil will have maximum e.m.f. induced in it when its plane lies parallel to the plane of the field even though it will have minimum flux linked with it. In general, the value of the induced e.m.f. is given by e = ωNΦ_m sinθ = E_m sinθ where θ is the angle between the axis of zero e.m.f. and the plane of the coil.

Here, f = 500/ 60 = 25/ 3 r.p.s ;

N = 100 ; B = 0.5 Wb/ m² ; A = (10 × 10) × 10⁻⁴ = 10⁻² m² .

∴ E_m= 2π f NBA = 2π (25/3) × 100 × 0.5 × 10⁻² = 26.2 V

(i) since θ = 0 ; sin θ = 0 ; therefore, e = 0.

(ii) Here, θ = 90° ; e = E_m sin 90º = 26.2 × 1 = 26.2 V

(iii) sin 45º = 1/√2 ; e = 26.2 × 1/√2 = 18.5 V

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