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Anit [1.1K]
3 years ago
6

Monday Homework Problem 10.6 A simple generator is constructed by rotating a flat coil in a uniform magnetic field. Suppose we r

otate 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
Physics
1 answer:
prohojiy [21]3 years ago
8 0

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

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