Respuesta :
Note: You didn't provide the diagram referred to but i can help you work around it without the diagram.
Answer:
1) Angular frequency of rotation = 5.24 rad/s
2) [tex]I_{max} = 0.24 A[/tex]
3) Magnitude of the magnetic flux = 0.091 wb
4) The induced current in the loop at t = 0.45, I₁ = 0.169 A
Explanation:
1) Angular frequency of rotation, w, is given by the formula:
[tex]\omega = \frac{2 \pi}{T}[/tex]
Period for one complete rotation, T = 1.2 seconds
[tex]\omega = \frac{2 \pi}{1.2} \\\omega = 5.24 rad/s[/tex]
b) The magnitude of the maximum current induced in the loop:
I = e/R
Where e = induced emf, and Resistance, R = 2.8 cm
e = NBAw
Since there is one complete rotation, N = 1
Magnetic field, B = 1.1 T
base, b = 28 cm = 0.28 m
height, h = 83 cm = 0.83 m
Area of the triangle, A = 0.5 * base * height
A = 0.5 * 0.28 * 0.83
A = 0.1162 m²
e = 1 * 1.1 * 0.1162 * 5.24
e = 0.6698 V
[tex]I_{max} = \frac{e sin 90}{R} \\I_{max} = \frac{0.6698 * 1}{2.8}\\I_{max} = 0.24 A[/tex]
3)
θ is the angle between the normal and the magnetic field
At t = 0, θ = 0
At t = 0.45 s, θ₁ = ?
[tex]\omega = \frac{\theta_{1} - \theta }{t_{1} -t}\\5.24 = \frac{\theta_{1} - 0 }{0.45 -0}\\5.24 = \frac{\theta_{1} }{0.45}\\\theta_1 = 5.24 * 0.45\\\theta_1 = 2.358 rad\\\theta_1 = 2.358 * 180/\pi \\\theta_1 = 135.1^{0}[/tex]
The magnetic flux is calculated as:
[tex]\phi = NBA cos \theta_{1} \\\phi = 1*1.1 * 0.1162 cos135.1\\\phi = -0.091 wb[/tex]
Magnitude of the magnetic flux = 0.091 wb
4)
Induced current, I = e/R
e = dΦ/dt
Φ = NBA cos(wt)
e = dΦ/dt = -NBAw sin (wt)
e = -(1 * 1.1 * 0.1162*5.24) sin(5.24 * 0.45*180/π)
e = -0.4736 V
I₁ = -0.4736/2.8
I₁ = -0.169 A
I₁ is defined to be positive due to the direction of flow
I₁ = 0.169 A
