Question

A conductor ABOCD moves along its bisector with a velocity 1 m/s through a perpendicular magnetic field of 1 wb/m², as shown in figure. If all the four sides are 1 m length each, then the induced emf between A and Din approx is ______V.

विकल्प

• 3.414

• 2.414

• 1.414

• 5.414

Answer 1

A conductor ABOCD moves along its bisector with a velocity 1 m/s through a perpendicular magnetic field of 1 wb/m², as shown in figure. If all the four sides are 1 m length each, then the induced emf between A and Din approx is 1.414 V.

Explanation:

V_A – V_D = V × B × (ℓ sin 45° + ℓ sin 45°)

1 × 1 × `sqrt2`

= 1. 41 V

sine rule `ℓ/(sin 90°) = 1/(sin 45°)`

⇒ ℓ = `sqrt2`

= 1.414