A 2.00 cm diameter, 11.0 cm long solenoid has 103 turns are has a 1.00 cm diamet

Question

A 2.00 cm diameter, 11.0 cm long solenoid has 103 turns are has a 1.00 cm diameter loop inside The loop has a resistance of 0.400 Ohm. What is the current in the loop at 0.0120 s? The the maximum and the minimum current in the graph have the same magnitude which is 1.60 A. In the figure provide, a magnet creates a magnetic field of strength 1.33 T. A hand crank generator consists of a semi circle with a radius of r = 6.90 cm as shown in the figure. The resistance in the bulb is 5.90 Ohm and the bulb dissipates 4.00 w of power: with what frequency will you have to turn the crank for the maximum current to fully light the bulb? Answer:

Explanation / Answer

Here ,

question 1 )

for the induced current in the loop

induced current in the loop = area of the loop *rate of change of magnentic field/R

induced current in the loop = pi * (0.01/2)^2 * 4pi *10^-7 * 103/0.11 * (1.60/.01)/0.40

induced current in the loop = 3.69 *10^-5 A

the induced current in the loop is 3.69 *10^-5 A

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