Question

Monday Homework Problem 10.6 A simple generator is constructed by rotating a flat coil in a uniform magnetic field. Suppose we rotate a flat coil of radius 2.0cm and 400 turns in the earth’s magnetic field 6 times per second so that the surface normal to the coil is parallel to the field once per revolution. The earth’s magnetic field is still about 4 × 10−5T. What is the maximum emf that would be measured acro

Answer 1

Our values are,

`B=1*10^(-5)T`

`N=400`

`r = 2.0cm = 0.02 m`

Number of rotations per second = 6

We begin calculating the cross-section Area, which is given by,

`A= \pi r^2`

`A= \pi (0.02)^2`

`A= 1.2566*10^(-3) m^2`

We can calculate the angular speed through the number of rotations per second,

`\omega = 6 rev`

`\omega = 6((2)/(pi))`

`\omega =37.699 rad/s`

With this velocity we can now calculate the maximum emf,

`E= BAN\omega`

`E = (4*10^(-5))(1.2566*10^(-3))(400)(37.699)`

`E= 7.579*10^(-4) V`

`E= 7.6*10^(-4) V`