Answer :
Answer:
emf=-10V
Explanation:
the emf can be calculated by using the expression
[tex]\epsilon=-N\frac{\Delta \Phi_B}{\Delta t}\\\Phi_B=BS[/tex]
where Ф is the magnetic flux and S is the area of the turns.
we can take that in one half of the period, we have the maximum and minimum of the flux. Hence
[tex]T=\frac{1}{f}=\frac{1}{10}=0.1s[/tex]
Thus, we have
[tex]\epsilon=-(100)\frac{(0.05T)(0.100m^2)}{\frac{0.1s}{2}}=10V[/tex]
hope this helps!!
To solve this problem we will apply the concepts related to the electric potential induced by a magnetic field. Mathematically it can be defined as the product between the number of loops, the magnetic field, the cross-sectional area and the angular velocity. Thus
[tex]\epsilon_{max} = NBA\omega[/tex]
Here,
N = Number of Loops
B = Magnetic Field
A = Cross-Sectional Area
[tex]\epsilon_{max} = (100)(0.05T)(0.1m^2)(10\frac{rev}{s} \frac{2\pi rad}{1rev})[/tex]
[tex]\epsilon_{max}=31.4 Volts[/tex]
Therefore the induced voltage is 31.4V