A sinusoidally oscillating current I ( t ) I(t) with an amplitude of 9.55 A 9.55 A and a frequency of 879 cycles per second 879 cycles per second is carried by a long, straight wire. A rectangular loop of copper wire with dimensions b = 72.2 cm b=72.2 cm by c = 32.5 cm c=32.5 cm is located a = 80.2 cm a=80.2 cm from the straight wire, and is coplanar with it. Calculate the average power P avg Pavg dissipated by the loop if its resistance is 64.3 Ω .

Answer: 1.4 seconds which is the third optionexplanation: current is defined as the flow of charge per timetherefore, the equation that relates current, charge and time can be expressed as follows: i = δq / δtwhere: i is the current given = 1.8 ampereδq is the charge given = 2.5 coulombδt is the time required for the flow of the given charges to give the 1.8 ampere.substitute with the givens in the above equation to get the time as follows: 1.8 = 2.5 / δtδt = 2.5 / 1.8 = 1.38888 which is approximately 1.4 secondshope this was !