Answer:
The magnitude of the induced emf in the coil is 5 volts.
Explanation:
Faraday's law states that an induced emf on a coil is:
[tex] \xi=-N\frac{d\varphi_{B}}{dt}[/tex] (1)
with [tex]\frac{d\varphi_{B}}{dt} [/tex] the change of magnetic flux on time and N the number of turns
Because magnetic field is perpendicular to the coil magnetic flux is [tex] \varphi_{B}=BA[/tex] with B the magnetic field and A the area of the coil, so the change on magnetic flux is [tex]\mid (3.0T)(0.5m)^2-(-1.0T)(0.5m)^2\mid [/tex] over a period of 2.0s is [tex]\frac{\mid (3.0T)(0.5m)^2-(-1.0T)(0.5m)^2\mid}{2s}=\frac{1Tm^2}{2s} [/tex].
Using that value on (1):
[tex]\xi=-(10)\frac{1}{2} [/tex]
Because we're interested on the magnitude the minus sign on the equation is not important here, so:
[tex]\mid \xi \mid =(10)\frac{1}{2}=5V [/tex]