- #1

physgrl

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## Homework Statement

An electric generator consists of n = 500 turns of wire formed into a rectangular loop with a length of 5 cm and width of 3 cm placed in a uniform magnetic field of 2.50 T. What is the maximum value of the emf produced when the loop is spun at f = 100 rpm about an axis perpendicular to B?

a. 19.6 V

b. 144 V

c. 95.3 V

d. 79.2 V

e. 60.3 V

## Homework Equations

ε=-NΔ∅

_{B}/Δt

∅

_{B}=BAcosβ

## The Attempt at a Solution

For a maximum emf there must be a maximum change in either B or A, but because A is constant B must be the variable changing. As the loop rotates B varies from 0T to 2.5T and it takes half a loop to do that.

ε=NAB/Δt

ε=500*(.03m*.05m)*(2.5T)/Δt

Δt

_{half a revolution}:

=>100rpm*1min/60s=1.67rev/sec

=> 1 revolution takes .6s

=> .5 revolutions take .3s

ε=500*(.03m*.05m)*(2.5T)/Δt

ε=500*(.03m*.05m)*(2.5T)/.3s

ε=6.25V

The answer I get is not in the options, and I don't see what I am doing wrong? Thanks!