Answer :
Answer:
The average induced emf is 50.88 V
Explanation:
Given;
number of turns, N = 80 turns
angle between the plane of the coil and magnetic field, θ = 43°
strength of magnetic field, B = 1.40 T
time to make 80 turns = 0.0700 s
[tex]Emf_{avg} = \frac{N \delta \phi}{dt} = N(\frac{BACos \theta {_f}-BACos \theta {_i}}{dt} )\\\\Emf_{avg} = \frac{NBA(Cos \theta {_f}-Cos \theta {_i})}{dt}[/tex]
where;
A is the area of the coil = 0.25 x 0.4 = 0.1 m²
If the plane makes an angle of 43° with a magnetic field to a position perpendicular to the field, then the initial angle θi = 90 - 43 = 47° and the final angle θf with the field = 0
[tex]Emf_{avg} = \frac{NBA(Cos \theta {_f}-Cos \theta {_i})}{dt}\\\\Emf_{avg} = \frac{80*1.4*0.1(Cos(0)-Cos(47))}{0.07} = 50.88 \ V[/tex]
Therefore, the average induced emf is 50.88 V
The average emf induced in the coil is; Emf = 50.88 V
We are given;
Number of turns; N = 80 turns
Length of rectangle; L = 40 cm = 0.4 m
Width of rectangle; W = 25 cm = 0.25 m
Magnetic field; B = 1.4 T
Change in rotation time; Δt = 0.07 s
Angle made by the plane of the coil with the magnetic field; θ = 43°
Magnetic field; B = 1.4 T
Now, formula for the average EMF Induced in the coil is given as;
Emf = NΔΦ/Δt
Where; ΔΦ = BAcos θ
where A is area = length x width = 0.25 * 0.4
A = 0.1 m²
Now, since the Angle made by the plane of the coil with the magnetic field was θ = 43°, then we can say that;
Initial angle; θ_i = 90° - 43° = 47°
final angle; θ_f = 0°
Thus; ΔΦ = BA(cos θ_f - cos θ_i)
∴Emf = (NBA(cos θ_f - cos θ_i))/Δt
∴Emf = [80 × 1.4 × 0.1(cos 0° - cos 47°)]/0.07
Emf = (80 × 1.4 × 0.1(1 - 0.6820))/0.07
Emf = 50.88 V
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