Answer:
0.0048 V
Explanation:
The side of the coil is
L = 20 cm = 0.20 m
So the area is
[tex]A=L^2 = (0.20)^2=0.04 m^2[/tex]
The magnetic flux through the coil at any instant t is
[tex]\Phi = BA cos (\omega t)[/tex]
where
B = 2.0 T is the magnetic flux density
A is the area of the coil
[tex]\omega[/tex] is the angular speed of the coil
t is the time
We know that the coil rotates at an angular speed of 10°/s, which converted into radians is
[tex]\omega = \frac{10 \cdot 2\pi}{360}=0.175 rad/s[/tex]
The magnitude of the induced emf in the loop is equal to the derivative of the magnetic flux, so:
[tex]\epsilon =- \frac{d\Phi}{dt}=-\frac{d}{dt}(BA cos(\omega t) = \omega B A sin(\omega t)[/tex] (1)
We want to know the magnitude of the emf when the angle is [tex]\theta=20^{\circ}[/tex], so substituting this value of the angle into [tex](\omega t)[/tex] in eq.(1), we find:
[tex]\epsilon = (0.175)(2.0)(0.04) sin (20^{\circ})=0.0048 V[/tex]
