Respuesta :
Answer:
a) 3.92 rad/s
b) 0.373 A
d) 0.018 A
Explanation:
a) The angular frequency of rotation is given by:
[tex]\omega=\frac{2\pi}{T}=\frac{2\pi}{(1.6s)}=3.92\frac{rad}{s}[/tex]
b) The maximum induced current in the loop is given by:
[tex]I_{max}=\frac{emf_{max}}{R}=\frac{AB\omega}{R}[/tex]
R: resistance
A: area of the triangle loop = bh/2 = (0.34m)(0.77m)/(2) = 0.1309m^2
B: magnitude of the magnetic field
[tex]I_{max}=\frac{(0.13m^2)(1.6T)(3.92rad/s)}{2.2\Omega}=0.373A[/tex]
d) For t = 0.45s you have:
[tex]I(t)=\frac{ABcos(\omega t)}{R}\\\\I(0.45)=\frac{(0.13m^2)(1.6T)cos(3.92rad/s \ (0.45s))}{2.2\Omega}=-0.018A[/tex]
But I1 is defined to be positive if it flows in the negative y-direction.
hence, I for t=0.45 s is 0.018A
(1) The angular frequency is 3.92 rad/s
(2) the value of I(max) = 0.373A
(3) Image is required
(4) The current at 0.45s is -0.018A
Induced current:
Given a conducting wire formed in the shape of a right triangle with:
Base b = 34 cm
Height h = 77 cm
Resistance R = 2.2 Ω
(1) Time period of rotation is T = 1.6s
The angular frequency is given by:
ω = 2π/T
ω = 2π/1.6
ω = 3.92 rad/s
(2) The magnetic field applied is B = 1.6T, perpendicular to the plane of the triangle.
the induced current is given by:
[tex]I_{ind}=\frac{EMF_{ind}}{R}[/tex]
where R is the resistance
[tex]I_{max}=\frac{EMF_{max}}{R}\\\\I_{max}=\frac{BA\omega}{R}\\\\I_{max}=\frac{1.6\times(1/2)\times34\times77\times3.92}{2.2}\\\\[/tex]
I(max) = 0.373A
(3) The image for the question is not attached.
(4) The instantaneous value of induced current at time t = 0.45s is given by:
[tex]I_{ind}=\frac{EMF_{ind}}{R}\\\\I(t)=\frac{BA(cos\omega t)}{R}\\\\I(t)=\frac{(1/2)\times34\times77\times1.6\timescos(3.92\imes0.45)}{2.2}[/tex]
I(t) = -0.018 A
the negative sign indicated that the current flows in the negative y-direction.
Learn more about induced current:
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