A circular copper disk of diameter D=10 cm rotates at frequency =1800 rev/min about an axis through its center and at right angles to the disk. A uniform magnetic field B of 10,000 gauss is perpendicular to the disk. What potential difference e develops between the axis of the disk and its rim?

Respuesta :

Answer:

e=2356.125 V

Explanation:

Given that

D= 10 cm  = 0.1 m

R= 0.05

B= 10,000

Angular speed ,ω = 1800 rev/min

Speed in rad/s given as

[tex]\omega =\dfrac{2\pi N}{60}[/tex]

[tex]\omega =\dfrac{2\pi \times 1800}{60}[/tex]

ω= 188.49 rad/s

The potential difference given as

e= B R v

v=average speed

[tex]v=\dfrac{\omega}{2}R[/tex]

[tex]e=\dfrac{B}{2}R^2 {\omega}[/tex]

By putting the values

[tex]e=\dfrac{10000}{2}0.05^2 \times {188.49}[/tex]

e=2356.125 V