Question

Determine the magnitude of emf (in v) induced between the axis of rotation and the rim of the disc, when the disc of radius 10 cm rotates with an angular velocity of 60 revolutions per second and placed in a magnetic field of 3 t acting parallel to the rotation of the disc.

Answer 1

Answer:

E = 0.565 V

Explanation:

As the disc rotates, any of its radii cuts the lines of force of magnetic field.

Area swept by radius vector during one revolution

= πr2

= π(10 cm)2

= 100π cm2

= π×10-2 m2

Area swept in one second,

A = (area swept in one revolution) × (Number of revolutions per second)

  = π×10-2×60

A = 0.6 πm2

Rate of change of magnetic flux

= dϕB/dt

= BA

= 0.3×0.6π

= 0.18π Wb

According to Faraday’s law, magnitude of induced e.m.f is,

E = dϕB/dt

Therefore, magnitude of e.m.f. is

E = 0.18 πV

Or,

E = 0.565 V

Answer 2

Answer:

E = 0.565 V

Explanation:

According to this question

As the disc rotates, any of its radii cut the lines of force of the magnetic field.

Area swept by radius vector during one revolution

= $πr^2$

= $π(10 cm)^{2}$

= $100π cm^{2}$

= $π10^{-2} m^{2}$

Area swept in one second,

A = (area swept in one revolution) × (Number of revolutions per second)

=$π\times 10^{-2}\times 60$

A = 0.6 πm2

Rate of change of magnetic flux

= $\frac{dϕB}{dt}$

= BA

= $(0.30)\times (0.6)π$

= 0.18π Wb

According to Faraday’s law, magnitude of induced e.m.f is,

E = $\frac{dϕB}{dt}$

Therefore, magnitude of e.m.f. is

E = 0.18 πV

Or,

E = 0.565 V