Answer:
the induced current is [tex]I = 3.2768*10^{-4} \ \ A[/tex]
Explanation:
Given that :
side of the square (l) = 16 cm = 0.16 m
Magnetic field B = 0.750 y² t
Resistance R = 0.500 Ω
Time t = 0.490 s
Let consider a small rectangular; whose length is [tex]l[/tex] and breath is [tex]dy[/tex]
Hence; to determine the magnetic flux through it small rectangular; we have:
[tex]d \phi = B.dA \\ \\ d \phi = B (i * dy) \\ \\ d \phi = (0.750 \ y^2 t ) *(l *dy) \\ \\ d \phi = ( 0.75*t*l) y^2 dy \\ \\ d\phi = (0.75 tl) y^2 dy[/tex]
Let calculate the total flux in the square loop
[tex]\phi = \int\limits^{y=l}_{y=0} B.dA \\ \\ \phi = \int\limits^{y=l}_{y=0} (0.75 \ t \ l ) y^2dy \\ \\ \phi = (0.75 \ t \ l )\frac{l^3}{3} \\ \\ \phi = (0.75 \ t )\frac{l^4}{3}[/tex]
Thus the total flux in the square loop is [tex]\phi = (0.75 \ t )\frac{l^4}{3}[/tex]
Now; going to the induced emf; let consider the Faraday's Law
[tex]V = \frac{d \phi}{dt}[/tex]
[tex]V = \frac{d }{dt}(\frac{0.75*l^4}{3})*t[/tex]
[tex]V = (\frac{0.75*0.16^4}{3})[/tex]
[tex]V = 1.6384*10^{-4} \ V[/tex]
Finally ; the induced current I is given by the expression;
[tex]I = \frac{V}{R} \\ \\ I = (\frac{1.6384*10^{-4}}{0.5})[/tex]
[tex]I = 3.2768*10^{-4} \ \ A[/tex]
Therefore; the induced current is [tex]I = 3.2768*10^{-4} \ \ A[/tex]
