In an ac dynamo, the peak value of emf is 60 V. The induced emf (in V) in the position when the armature makes an angle of 30° with the magnetic field perpendicular to the coil, will be

Question

In an ac dynamo, the peak value of emf is 60 V. The induced emf (in V) in the position when the armature makes an angle of 30° with the magnetic field perpendicular to the coil, will be (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

ε=ε0 sin ω t=ε0 sin θ=60 sin 30°=30 V

Q. In an ac dynamo, the peak value of emf is $60 V$ . The induced emf (in $V$ ) in the position when the armature makes an angle of $3 0^{\circ}$ with the magnetic field perpendicular to the coil, will be

$ε = ε_{0} sin ω t = ε_{0} sin θ = 60 sin 3 0^{\circ} = 30 V$

Q. In an ac dynamo, the peak value of emf is 60V. The induced emf (in V) in the position when the armature makes an angle of 30∘ with the magnetic field perpendicular to the coil, will be

ε=ε0​sinωt=ε0​sinθ=60sin30∘=30V

Q. In an ac dynamo, the peak value of emf is $60 V$ . The induced emf (in $V$ ) in the position when the armature makes an angle of $3 0^{\circ}$ with the magnetic field perpendicular to the coil, will be

$ε = ε_{0} sin ω t = ε_{0} sin θ = 60 sin 3 0^{\circ} = 30 V$