The flux density B in a region between the pole faces of a horseshoe magnet 0.5 Wb m⁻² directed vertically downwards. Find the emf induced in a straight wire 5 cm long, perpendicular to B when it is moved in a direction at an angle of 60° with the horizontal with a speed of 100 cm s⁻¹. (Ans; 1.25 𝐱 10⁻² V)
Data Given:
Flux density = B = 0.5 Wb m⁻²
Length of wire = L = 5 cm = 0.05 m
Speed of the wire = v = 100 cm s⁻¹ = 1 m s⁻¹
As given speed v make an angle of 60° with horizontal. But we need an angle between the v and B, Thus,
Angle = θ = 90° - 60° = 30°
To Find:
Induced emf = Ɛ = ?
Solution:
Using the following relationsƐ = V B L sin θ
By putting values
Ɛ = 1 m s⁻¹ x 0.5 Wb m⁻² x 0.05 m x sin 30°
Ɛ = 1 m s⁻¹ x 0.5 Wb m⁻² x 0.05 m x sin 30°
Ɛ = 0.025 Wb s⁻¹ x 0.5
Ɛ = 0.0125 A
Ɛ = 1.25 x 10⁻² A ----------Ans.
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