Answer:
Induced current,[tex]I=2.87\times 10^{-3}\ A[/tex]
Explanation:
Given that,
Area of cross section of the wire, [tex]A=200\ cm^2=0.02\ m^2[/tex]
Time, t = 2.2 s
Initial magnetic field, [tex]B_i=0.2\ T[/tex]
Final magnetic field, [tex]B_f=2.8\ T[/tex]
Resistance of the coil, R = 8 ohms
The expression for the induced emf is given by :
[tex]\epsilon=-\dfrac{d\phi}{dt}[/tex]
[tex]\phi[/tex] = magnetic flux
[tex]\epsilon=-\dfrac{d(BA)}{dt}[/tex]
[tex]\epsilon=A\dfrac{d(B)}{dt}[/tex]
[tex]\epsilon=A\dfrac{B_f-B_i}{t}[/tex]
[tex]\epsilon=0.02\times \dfrac{2.8-0.2}{2.2}[/tex]
[tex]\epsilon=-0.023\ volts[/tex]
So, the induced emf in the loop is 0.023 volts. The induced current can be calculated using Ohm's law as :
[tex]\epsilon= IR[/tex]
[tex]I=\dfrac{\epsilon}{R}[/tex]
[tex]I=\dfrac{0.023}{8}[/tex]
[tex]I=2.87\times 10^{-3}\ A[/tex]
So, the magnitude of the induced current in the loop of wire is [tex]2.87\times 10^{-3}\ A[/tex]. Hence, this is the required solution.
