Question

An electric generator is constructed using a permanent magnet with magnetic field magnitude, 10mT. A circular loop of wire of radius, 50cm is rotated at a rate of 5 rev/s. (a) Write an expression for the power output (as a function of time) by a lightbulb with resistance 15kΩ which is hooked up to this generator.

(b) Would the power output increase or decrease if we decrease the resistance of the lightbulb?

Answer 1

(a). The power output by a light bulb is
`4.00*10^(-6)\ watt`.

(b). The power output is increase.

Step-by-step explanation:

Given that,

Magnetic field = 10 mT

Radius = 50 cm

Angular velocity = 5 rev/s

Resistance = 15 kΩ

We need to calculate the area

Using formula of area

`A=\pi r^2`

`A=\pi*(50*10^(-2))^2`

`A=0.78\ m^2`

(a). We need to calculate the power output by a light bulb

Using formula of power

`P=(E^2)/(R)`

`P=(BA\omega)/(R)`

Put the value into the formula

`P=((10*10^(-3)*0.78*5*2\pi)^2)/(15*10^(3))`

`P=0.00000400\ watt`

`P=4.00*10^(-6)\ watt`

(b). We need to calculate the power output increase or decrease if we decrease the resistance of the light-bulb

If we decrease the resistance of the light bulb then the power output will be increase. Because the power output is inversely proportional to the resistance.

Hence, (a). The power output by a light bulb is
`4.00*10^(-6)\ watt`.

(b). The power output will be increase.