A current with sinusoidal time-dependence, whose peak value is 3.55 A and frequency is 463 cycles per second, passes through a long straight wire. A rectangular 72.2 cm × 32.5 cm loop of copper wire is located 80.2 cm from the wire and is coplanar with it. Calculate the average power dissipated by the loop, if its resistance is 34.3 ohms.
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To calculate the average power dissipated by the loop, we can use the formula:
Average Power (P) = (1/2) * I_m^2 * R,
where I_m is the peak current in the sinusoidal waveform and R is the resistance of the loop.
To find the average power, we need to find the peak current (I_m). We can do this by using the formula:
I_m = I_peak * √2,
where I_peak is the peak value of the current.
Given:
I_peak = 3.55 A (peak value of the current)
Frequency = 463 cycles/second
The time period (T) can be calculated as:
T = 1 / frequency = 1 / 463
Now, we can calculate the peak current (I_m):
I_m = I_peak * √2
Next, calculate the root-mean-square (RMS) current (Irms):
Irms = I_m / √2
Now, we have the RMS current (Irms), which can be used to calculate the average power dissipated by the loop:
P = (1/2) * Irms^2 * R
Given:
R = 34.3 ohms (resistance of the loop)
Now, let's plug in the values and solve for the average power (P).