Question

A SQUARE induction coil with a side length 3 cm and 400 windings, is placed perpendicularly in a uniform magnetic field and then rotated through an angle of 45⁰ in 0,08 s. C E S An emf of 7 V is induced in the coil. 3.1 Calculate the change in the magnetic flux. 3.2 Calculate the magnitude of the magnetic field.​

Answer 1

The result will be E1=E2>E4>E3

When the plane of the armature of an a.c. dynamo is parallel to the lines of force of the magnetic field, the amplitude of the induced e.m.f. is at its greatest. Additionally, the rate of change of the magnetic flux associated with the armature is at its greatest. As a result, in this orientation, the e.m.f. induced in the armature is greatest.

The EMF's amplitude is equal to Nr2B. N is the number of turns, and r is the square's half-side.

In example 3, the axis is perpendicular to the magnetic field, hence the flux is zero and the amplitude is zero as a result.

Case 4 has an axis that falls between types 1, 2, and 3.

Here, we must consider the angular velocity's x and y components.

Component in the x direction will be zero.

Additionally, the new angular velocity will be 2w because it is exactly halfway between x and y.

So now put these numbers in actual formula. The result will be E1=E2>E4>E3

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