Question

**9**. A magnetic field is perpendicular to the
plane of a single-turn circular coil. The magnitude of the field is
changing, so that an emf of 0.77 V and a current of 2.7 A are
induced in the coil. The wire is then re-formed into a single-turn
square coil, which is used in the same magnetic field (again
perpendicular to the plane of the coil and with a magnitude
changing at the same rate). What emf is induced in the square
coil?

V

Answer #1

Solution:-

Since V = 0.77V and I = 2.7A the resistance is V/I = 0.77/2.7 =
0.285Ω

The induced emf = dB/dt *A therefore dB/dt = Emf/A

Since dB/dt is constant then Emf/A(circle) = Emf/A square

So Emf (square)/Emf (circle) = A square / A circle

Area of a circle = π*r^2

The circumference is 2πr

The perimeter of the square is 2πr which makes each side πr/2
This means the area of the square is π^2*r^2/4

So A square/Acircle = π^2*r^2/4/ π*r^2 = π/4 = 0.785

So Emf square = Emf circle *0.785 = 0.77*0.785 =
**0.604V**

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