Respuesta :
Answer:
The induced emf is [tex]|\epsilon|=0.0261 V[/tex]
Explanation:
From the question we are told that
The number of turn is [tex]N = 100[/tex]
The diameter of the coil is [tex]d = 2.0 cm = 2.0 *10^{-2} m[/tex]
The uniform magnetic at initial is [tex]B_i = 0.50 T[/tex]
The uniform magnetic at initial is [tex]B_f = 1.50 T[/tex]
The time taken is [tex]t = 0.60s[/tex]
The angle the magnetic field makes with vertical is [tex]\theta = 60^o[/tex]
Generally induced emf is mathematically represented as
[tex]\epsilon = -N \frac{d \o}{dt}[/tex]
where [tex]d \o[/tex] is the change magnetic flux
Magnetic flux is mathematically represented as
[tex]\O = \= B \cdot \= A[/tex]
[tex]= BA cos \theta[/tex]
Substituting this above
[tex]\epsilon = -N \frac{d (BA cos \theta)}{dt}[/tex]
[tex]\epsilon = -N A \frac{d (B cos \theta)}{dt}[/tex]
Where B is the magnetic field and A is the area which is mathematically evaluated as
[tex]A = \frac{\pi d^2}{4}[/tex]
Substituting values
[tex]A = \frac{\pi (2.0 *10^{-2})^2}{4}[/tex]
[tex]A= 3.142*10^{-4}m^2[/tex]
From the equation of emf
[tex]\epsilon = -N A \frac{d (B cos \theta)}{dt}[/tex]
dB = [tex]B_2 -B_1[/tex]
So
[tex]|\epsilon| = N A \frac{ (B_2 -B_1 cos \theta)}{dt}[/tex]
substituting values
[tex]|\epsilon| = 100(3.142*10^{-4}) \frac{ (1.50 -0.50 cos(60))}{0.60}[/tex]
[tex]|\epsilon|=0.0261 V[/tex]
The EMF induced in the coil as the magnetic field changes with time is 0.0262V which is explained below.
Induced EMF:
The EMF induced in the coil is given by:
E = -NΔФ/Δt
initially, the flux through the coil is:
Ф = BAcosθ
here it is given that θ = 60°
A is the area of the coil = πd²/4
A = 3.14 × (0.02)²/4 m²
A = 3.14×10⁻⁴ m²
and, B is the initial magnetic field = 0.50T
So,
Ф = 0.5×3.14×10⁻⁴×cos60
similarly, final flux:
Ф' = 1.5×3.14×10⁻⁴×cos60
so, the change in flux is:
ΔФ = Ф'-Ф = 1×3.14×10⁻⁴×cos60
ΔФ = 1.57×10⁻⁴ Wb
So, the magnitude of EMF generated is :
| E | = NΔФ/Δt
| E | = 100×1.57×10⁻⁴/0.6
| E | = 0.0262V
Learn more about induced EMF:
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