Question

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 0.23 V and a current of 2.4 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 (a) emf and (b) current are induced in the square coil?

Answer #1

V = 0.23V and I = 2.4 A

the resistance is V/I = 0.23/2.4 = 0.095 Ω

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

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

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

so A circle = π*r^2 the circumference is 2πr so 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 = 0.7854

a) So Emf square = Emf circle *0.7854 = 0.23*0.7854 = 0.1806
V

b) I = V/R = 0.1806 V/0.095 =1.901 A

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