Answer:
the thermal energy generated in the loop = [tex]6.64*10^{-4} \ W[/tex]
Explanation:
Given that;
The length of the copper wire L = 0.614 m
Radius of the loop r = [tex]\frac{L}{2 \pi}[/tex]
r = [tex]\frac{0.614}{2 \pi}[/tex]
r = 0.0977 m
However , the area of the loop is :
[tex]A_L = \pi r^2[/tex]
[tex]A_L = \pi (0.0977)^2[/tex]
[tex]A_L = 0.02999 \ m^2[/tex]
Change in the magnetic field is [tex]\frac{dB}{dt}= 0.0914 \ T/s[/tex]
Then the induced emf e = [tex]A_L \frac{dB}{dt}[/tex]
e = [tex]0.02999 * 0.0914[/tex]
e = 2.74 × 10⁻³ V
resistivity of the copper wire [tex]\rho = 1.69* 10^{-8}[/tex] 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 = [tex]\frac {\rho L}{\pi r^2}[/tex]
R = [tex]\frac{(1.69*10^{-8})(0.614)}{ \pi (0.54*10^{-3})^2}[/tex]
R = 1.13× 10⁻² Ω
Finally, the thermal energy generated in the loop (i.e the power) = [tex]\frac{e^2}{R}[/tex]
= [tex]\frac{(2.74*10^{-3})^2}{1.13*10^{-2}}[/tex]
= [tex]6.64*10^{-4} \ W[/tex]
