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 relations

Ɛ = 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 = 0.025 Wb s⁻¹  ｘ 0.5

Ɛ = 0.0125 A 

Ɛ = 1.25 ｘ 10⁻² A ----------Ans.



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