Respuesta :
Answer:
0.025V + (0.000218V/s³) t³
Explanation:
Parameters given:
Radius of coil, r = 3.85 cm = 0.0385 m
Number of turns, N = 450
Magnetic field, B = ( 1.20×10^(−2) T/s )t + (2.60×10^(−5) T/s4 )t^4.
The magnitude of Induced EMF is given as:
E = N * A * dB/dt
Where A is the area of the coil
First, we differentiate the magnetic field with respect to time:
dB/dt = 0.012 + 0.000104t³
Therefore, EMF will be:
E = 450 * 3.142 * (0.012 + 0.000104t³)
E = 2.096(0.012 + 0.000104t³)
E = 0.025V + (0.000218V/s³)t³
Answer:
The magnitude of the induced emf is [tex]E=0.02511V+(0.000217\frac{V}{s^{3} } )t^{3}[/tex]
Explanation:
The area is:
[tex]A=\pi r^{2}[/tex]
r = 3.85 cm = 0.0385 m
[tex]A=\pi *0.0385^{2} =0.00465m^{2}[/tex]
The induce emf is:
[tex]E=NA\frac{dB}{dt}[/tex]
Replacing
[tex]E=450*0.00465*((\frac{d(1.2x10^{-2})t }{dt} )+(\frac{d(2.6x10^{-5})t^{4} }{dt} )\\E=2.0925*(1.2x10^{-2} +0.000104t^{3} )\\E=0.02511V+(0.000217\frac{V}{s^{3} } )t^{3}[/tex]