An application of electromagnetic induction occurs when you shake a Faraday flas

Question

An application of electromagnetic induction occurs when you shake a Faraday flashlight. You and your Team Members decide to design your own Faraday flashlight. In your teams design, a magnet is inserted in the handle and the magnet is free to slide back and forth through a loop with 50,000 turns. This energy is stored in a whopper of a capacitor (1.0 F) and can be retrieved to power a 0.50-W LED bulb. Your team measures the area of the loop to be 3.0x10-4 m2. When the magnet is at its farthest position, the magnetic field is measured to be 1.0 T. To convert the induced ac emf to dc (so the capacitor is charging when the magnet moves either way), a voltage rectifier is used. If the circuit resistance is measured to be 500 , estimate the number of shakes (one motion of the magnet back and forth) necessary to produce enough energy for 5.0 minutes of operation.

Explanation / Answer

N = 50000,Capacitance, C = 1 F,Power, P = 0.5 W,Area, A = 3*10^-4 m2, R = 500 ohm

Now,Flux through the loop, Q = NBA

Hence induced emf, E = dQ/dt = N*A*dB/dt

= 50000*3*10^-4*dB/dt = 15*dB/dt

Hence induced current, I = E/R = (15*dB/dt)/500

= 0.03*dB/dt

again let the number of shakes necessary in 5 minutes(or 5*60 = 300 seconds) be taken as x

then number of shakes in 1 second = x/300

its observed that number of shakes in 1 s = rate of change of magnetic field = dB/dt

x/300 = dB/dt

I = 0.03*(x/300)

the total charge stored in capacitor, Q = I*t

where t = 5 mins = 300 s

So, Q = 0.03*(x/300)*300 = 0.03*x

Power , P = Energy/time

energy stored,U = P*t = 0.5*300 = 150 J

hence Energy stored in capacitor, U = Q^2/(2C) = (0.03*x)^2/(2*1) = 4.5*10^-4*x^2

4.5*10^-4*x^2 = 150

calculating x

x = 578

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