Question

A 16.0 cm × 16.0 cm square loop of wire lies in the xy-plane with its bottom edge on the x-axis. The resistance of the loop is 0.500 Ω . A magnetic field parallel to the z-axis is given by B = 0.750 y2t, where B is in tesla, y is in meters, and t is in seconds. What is the size of the induced current in the loop at t = 0.490 s ? Express your answer with the appropriate units.

Answer 1

Answer:

the induced current is $I = 3.2768*10^{-4} \ \ A$

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 $l$ and breath is $dy$

Hence; to determine the magnetic flux through it small rectangular; we have:

$d \phi = B.dA \\ \\ d \phi = B (i * dy) \\ \\ d \phi = (0.750 \ y^2 t ) *(l *dy) \\ \\ d \phi = ( 0.75*t*l) y^2 dy \\ \\ d\phi = (0.75 tl) y^2 dy$

Let calculate the total flux in the square loop

$\phi = \int\limits^{y=l}_{y=0} B.dA \\ \\ \phi = \int\limits^{y=l}_{y=0} (0.75 \ t \ l ) y^2dy \\ \\ \phi = (0.75 \ t \ l )\frac{l^3}{3} \\ \\ \phi = (0.75 \ t )\frac{l^4}{3}$

Thus the total flux in the square loop is $\phi = (0.75 \ t )\frac{l^4}{3}$

Now; going to the induced emf; let consider the Faraday's Law

$V = \frac{d \phi}{dt}$

$V = \frac{d }{dt}(\frac{0.75*l^4}{3})*t$

$V = (\frac{0.75*0.16^4}{3})$

$V = 1.6384*10^{-4} \ V$

Finally ; the induced current I is given by the expression;

$I = \frac{V}{R} \\ \\ I = (\frac{1.6384*10^{-4}}{0.5})$

$I = 3.2768*10^{-4} \ \ A$

Therefore; the induced current is $I = 3.2768*10^{-4} \ \ A$