Answer:
a) 43.20V
b) 2.71W/s
c) 40.25s
d) 7.77Nm
Explanation:
(a) The emf of a rotating coil with N turns is given by:
[tex]emf=NBA\omega sin(\omega t)[/tex]
N: turns
B: magnitude of the magnetic field
A: area
w: angular velocity
the emf max is given by:
[tex]emf_{max}=NBA\omega=(50)(1.80T)(0.200m*0.100m)(24.0rad/s)\\\\emf_{max}=43.20V[/tex]
(b) the maximum rate of change of the magnetic flux is given by:
[tex]\frac{d\Phi_B}{dt}=\frac{d(A\cdot B)}{dt}=\frac{d}{dt}(ABcos\omega t)=AB\omega sin(\omega t)\\\\\frac{d\Phi_B}{dt}_{max}=(\pi(0.200*0.100))(1.80T)(24.0rad/s)=2.71\frac{W}{s}[/tex]
(c) [tex]emf(t=0.050s)=(50)(1.80T)(0.200m*0.100m)(24rad/s)sin(24.0rad/s(0.050s))\\\\emf(t=0.050s)=40.26V[/tex]
(d) The torque is given by:
[tex]\tau=NABIsin\theta\\\\NAB\omega=emf_{max}\\\\\tau=\frac{emf_{max}}{\omega}\frac{emf_{max}}{R}\\\\\tau=\frac{(43.20V)^2}{(24.0rad/s)(10.0\Omega)}=7.77Nm[/tex]