Answer:
a) emf = 0.01804 V
b) I = 0.03 A
Explanation:
a) The emf is calculated by using the following formula:
[tex]|emf|=|\frac{d\Phi_B}{dt}|=|\frac{d(A\cdot B)}{dt}|[/tex] [tex]=A|\frac{dB}{dt}|[/tex]
A: area of the loop = 0.0820m^2
B: magnitude of the magnetic field
dB/dt: change of the magnetic field, in time: 0.220 T/s
Where ФB is the magnetic flux, the surface vector and magnetic vector are perpendicular between them, and the area A is constant.
You replace the values of A and dB/dt in the equation (1):
[tex]|emf|=(0.082m^2)(0.220T/s)=0.01804V[/tex]
b) The current in the loop is:
[tex]I=\frac{emf}{R}[/tex]
R: resistance of the loop = 0.600Ω
[tex]I=\frac{0.01804V}{0.600\Omega}=0.03A=30mA[/tex]
a. The emf induced in this loop is 18.04mV.
b. The current induced in the loop is 30.06mA.
a. We know that,
[tex]flux(\phi)=B*A[/tex]
Where B is magnetic field and A is the area.
[tex]emf=\frac{d\phi}{dt}=A*\frac{dB}{dt}[/tex]
Given that, Area , [tex]A=0.0820m^{2},B=3.80T,\frac{dB}{dt}=0.220T/s[/tex]
Substituting all values in above equation.
[tex]emf=0.0820*0.220=0.01804V=18.04mV[/tex]
b. Resistance, [tex]R=0.600ohm[/tex]
Current induced in the loop is,
[tex]I=\frac{emf}{R}=18.04/0.6=30.06mA[/tex]
Hence, the emf induced in this loop is 18.04mV.
The current induced in the loop is 30.06mA.
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