Answer:
0.0129 V
Explanation:
The magnitude of the induced emf in the circuit is given by:
[tex]\epsilon = \frac{\Delta \Phi}{\Delta t}[/tex]
where
[tex]\Delta \Phi[/tex] is the change in magnetic flux through the coil
[tex]\Delta t[/tex] is the time interval
To find the change in magnetic flux, we need to find the initial flux and the final flux.
The area of the coil is
[tex]A=\pi r^2 = \pi (0.35 m)^2=0.385 m^2[/tex]
The initial magnetic field is
[tex]B_i = 0.50 T[/tex]
so the initial flux is
[tex]\Phi_i = B_i A = (0.50 T)(0.385 m^2)=0.193 Wb[/tex]
While the final flux is zero, since the coil is completely out of the magnetic field:
[tex]\Phi_f = 0[/tex]
so the magnitude of the change in flux is
[tex]\Delta \Phi = |\Phi_f - \Phi_i|=|0-0.193 Wb|=0.193 Wb[/tex]
While the time interval is
[tex]\Delta t = 15 s[/tex]
so the induced emf is
[tex]\epsilon = \frac{0.193 Wb}{15 s}=0.0129 V[/tex]
