Answer :
The magnitude of the current induced in the wire is 1.2×10^{-8} and direction of the current induced in the wire is counter clockwise.
In the given question,
A rectangle (W = 20 cm, H = 30 cm) and has a resistance of(R) = 5.0 mΩ.
B(s) = 4.0 μT and B(b) = -2.5
t(s) = 2.0 s
From the question
dB(1)/dt = B(s)/t(s)
dB(1)/dt = 4 μT/2
As μ = 10^{-6}
dB(1)/dt = 2×10^{-6} T/s
dB(2)/dt = 1/2dB(1)/dt
dB(2)/dt = 2×10^{-6} /2
dB(2)/dt = 1×10^{-6} T/s
dB(3)/dt = B(b)dB(1)/dt
dB(3)/dt = -2.5×2×10^{-6} T/s
dB(3)/dt = -5×10^{-6} T/s
Now the change in magnetic field is,
dB/dt = dB(1)/dt + dB(2)/dt + dB(3)/dt
dB/dt = 2×10^{-6} + 1×10^{-6}/2 - 5×10^{-6}
dB/dt = - 2×10^{-6}
Now finding the area
Area = 1/2 wh
Firstly converting the w and h in m
So w = 0.20 m, h = 0.30 m
Area = 1/2 *0.20*0.30
Area = 0.03 square m
The induced EMF is
EMF = -AdB/dt
EMF = -(0.03)*(- 2×10^{-6})
EMF = 0.06×10^{-6}
The magnitude of the current induced in the wire is
I = EMF/R
I = 0.06×10^{-6}/5
I = 0.012×10^{-6}
I = 1.2×10^{-8}
Now, the direction of the current induced in the wire is counter clockwise.
To learn more about magnitude of the current link is here
brainly.com/question/14526577
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The right question is:
Figure (a) shows a wire that forms a rectangle (W = 20 cm, H = 30 cm) and has a resistance of 5.0 mΩ. Its interior is split into three equal areas, with magnetic fields B(1), B(2), and B(3). The fields are uniform within each region and directly out of or into the page as indicated. Figure (b) gives the change in the z components B(z) of the three fields with time t; the vertical axis scale is set by B(s) = 4.0 μT and B(b) = -2.5 B(s) and the horizontal axis scale is set by t(s) = 2.0 s. What are the magnitude and direction of the current induced in the wire?
