Answer:
Induced emf in the loop is 0.02208 volt.
Induced current in the loop is 0.0368 A.
Explanation:
Given that,
Area of the single loop, [tex]A=0.092\ m^2[/tex]
The initial value of uniform magnetic field, B = 3.8 T
The magnetic field is decreasing at a constant rate, [tex]\dfrac{dB}{dt}=0.24\ T/s[/tex]
(a) The induced emf in the loop is given by the rate of change of magnetic flux.
[tex]\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=A\times \dfrac{dB}{dt}\\\\\epsilon=0.092\times 0.24\\\\\epsilon=0.02208\ V[/tex]
(b) Resistance of the loop is 0.6 ohms. Let I is the current induced in the loop. Using Ohm's law :
[tex]\epsilon=IR\\\\I=\dfrac{\epsilon}{R}\\\\I=\dfrac{0.02208}{0.6}\\\\I=0.0368\ A[/tex]
Hence, this is the required solution.
