Answer:
the thermal energy generated in the loop =
Explanation:
Given that;
The length of the copper wire L = 0.614 m
Radius of the loop r =
r =
r = 0.0977 m
However , the area of the loop is :
Change in the magnetic field is
Then the induced emf e =
e =
e = 2.74 × 10⁻³ V
resistivity of the copper wire Ω m
diameter of the wire = 1.08 mm
radius of the wire = 0.54 mm = 0.54 × 10⁻³ m
Thus, the resistance of the wire R =
R =
R = 1.13× 10⁻² Ω
Finally, the thermal energy generated in the loop (i.e the power) =
=
=