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,

$\omega = 6 rev$

$\omega = 6(\frac{2}{pi})$

$\omega =37.699 rad/s$

With this velocity we can now calculate the maximum emf,

$E= BAN\omega$

$E = (4*10^{-5})(1.2566*10^{-3})(400)(37.699)$

$E= 7.579*10^{-4} V$

$E= 7.6*10^{-4} V$