Respuesta :
Answer:
0.07 J
Explanation:
Parameters given:
Radius of loop, r = 13 cm = 0.13 m
Change in magnetic field, B = 0.86 T
Time taken, t = 0.67 s
Resistance per unit length of wire, R/L = [tex]4.4 * 10^{-2} ohms/m[/tex]
Resistance in the entire length of wire, R = R/L * L (where L = circumference of wire)
R = [tex]4.4 * 10^{-2} * 2\pi r = 4.4 * 10^{-2} * 2\pi * 0.13[/tex]
R = 0.036 ohms
Electrical energy is given as the product of Power and Time;
E = P * t
Electrical power, P, is given as:
[tex]P = \frac{V^2}{R}[/tex]
where V = EMF/Voltage
The average EMF induced in a coil is given as:
[tex]V = \frac{-BA}{t}[/tex]
where A = Area of coil = [tex]\pi r^2[/tex] = [tex]0.13^2\pi = 0.0531 m^2[/tex]
Therefore, Power becomes:
[tex]P = \frac{(\frac{-BA}{t})^2 }{R}\\ \\\\P = \frac{(BA)^2}{Rt^2}[/tex]
Then, Electrical energy becomes:
[tex]E = \frac{(BA)^2}{Rt^2} * t\\\\\\E = \frac{(BA)^2}{Rt}[/tex]
[tex]E = \frac{(0.86 * 0.053)^2}{(4.4 *10^{-2} * 0.67)}\\ \\\\E = 0.07 J[/tex]
The average electrical energy dissipated in the resistance of the wire is 0.07 J
