A 70-turn coil has a diameter of 11 cm. Find the magnitude of the emf induced in the coil (in V) if it is placed in a spatially uniform magnetic field of magnitude 0.70 T so that the face of the coil makes the following angles with the magnetic field, and the magnetic field is reduced to zero uniformly in 0.2 s.

This question is incomplete, the complete question is;

A 70-turn coil has a diameter of 11 cm. Find the magnitude of the emf induced in the coil (in V) if it is placed in a spatially uniform magnetic field of magnitude 0.70 T so that the face of the coil makes the following angles with the magnetic field, and the magnetic field is reduced to zero uniformly in 0.2 s. a) θ = 30° V b) θ = 60° V c) θ = 90° V

the magnitude of the emf induced in the coil are;

a)- For θ = 30°, e = 1.16 V

b)- For θ = 60°, e = 2.01 V

c)- For θ = 90°, e = 2.33 V

Explanation:

Given the data in the question;

number of turns N = 70

Diameter of coil D = 11 cm

Radius r = D/2 = 11/2 = 5.5 cm = 0.055 m

magnitude of magnetic ΔB = 0.7T

Δt = 0.2 seconds

Now,

a)

For θ = 30°,

Angle of with area of vector θ' = 90° - 30° = 60°

so

emf  e = N( Δ∅ / Δt ) = N( ΔBAcosθ  / Δt )

hence

e =  NAcosθ'(ΔB / Δt )

where A is area ( πr² )

so we substitute

e =  70 × πr² × cos(60°) × ( 0.7 / 0.2 )

e =  70 × π(0.055)² × cos(60°) × ( 0.7 / 0.2 )

e = 1.16 V

b)

For θ = 60°,

Angle of with area of vector θ' = 90° - 60° = 30°

so

e =  NAcosθ'(ΔB / Δt )

we substitute

e =  70 × πr² × cos(30°) × ( 0.7 / 0.2 )

e =  70 × π(0.055)² × cos(30°) × ( 0.7 / 0.2 )

e = 2.01 V

c)

For θ = 90°,

Angle of with area of vector θ' = 90° - 90° = 0°

so

e =  NAcosθ'(ΔB / Δt )

we substitute

e =  70 × πr² × cos(0°) × ( 0.7 / 0.2 )

e =  70 × π(0.055)² × cos(30°) × ( 0.7 / 0.2 )

e = 2.33 V

Therefore, the magnitude of the emf induced in the coil are;

a)- For θ = 30°, e = 1.16 V

b)- For θ = 60°, e = 2.01 V

c)- For θ = 90°, e = 2.33 V

no sign of dia of fuel cell