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You want to build a generator to light a bulb. You have a fairly large permanent magnet, some wire, and a lightbulb. You may build it as follows. Wind the wire for about 200 turns into a coil of 15 cm radius. Tape up the coil and place it in the region between the poles of the magnet that produces a 0.10 T magnetic field. Connect a hand crank to the coil and rotate it at a rotational speed ω.

What is the required rotational speed ω to generate a peak voltage of 6 V?

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oladayoademosu

The required rotational speed ω to generate a peak voltage of 6 V is 4.24 rad/s

The induced peak voltage

The induced peak voltage in the generator is given by ε = NABω where

  • N = number of turns of coil = 200,
  • A = area of coil = πr² where
  • r = radius of coil = 15 cm = 0.15 m,
  • B = magnetic field strength = 0.10 T and
  • ω = rotational speed of coil

Rotational speed

Making ω subject of the formula, we have

ω = ε/NAB

ω = ε/Nπr²B

Substituting the values of the variables into the equation, we have, given that ε = 6 V

ω = ε/Nπr²B

ω = 6 V/(200 turns × π(0.15 m)² × 0.10 T)

ω = 6 V/(200 turns × 0.0225π m² × 0.10 T)

ω = 6 V/(200 turns × 0.0225π m² × 0.10 T)

ω = 6 V/0.45π turns-m²T

ω = 6 V/1.4137 turns-m²T

ω = 4.244 rad/s

ω ≅ 4.24 rad/s

The required rotational speed ω to generate a peak voltage of 6 V is 4.24 rad/s

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