A length of 20-gauge copper wire (of diameter 0.8118 mm) is formed into a circular loop with a radius of 25.0 cm. A magnetic field perpendicular to the plane of the loop increases from zero to 10.0 mT in 0.34 s. Find the average electrical power dissipated in the process

The average electrical power dissipated in the process is 0.653 mW

Explanation:

Given;

gauge of copper wire, 20 gauge

resistivity from chart,

radius of the circular loop, r = 25 cm

magnetic field strength, B = 10 .0 mT

time, t = 0.34 s

Length of the wire, L = 2πr = 2 x π x 0.25 = 1.571 m

Area of the wire, A = πR² ⇒ R = D/2 = 0.8118 mm/ 2 = 0.4059 mm

= π(0.4059 x 10⁻³)² = 0.5177 x 10⁻⁶ m²

The resistance of the wire is given by;

Now, determine the electric potential;

The average power is given by;

P = V²/R

P = (0.00577²) / (5.098 x 10⁻²)

P = 6.53 x 10⁻⁴ W

P = 0.653 mW

Therefore, the average electrical power dissipated in the process is 0.653 mW

at positions a, g kinetic energy is maximum.

at position d, potential energy is maximum.

i think it would be 5