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Grade: 12th pass
A copper disc 20 cm in diameter and turns 1000 times to form a coil. The disc is placed in a magnetic field of induction 0.27 acting parallel to the axis of rotation of the disc. Calculate the magnitude between the emf induced between the axis of rotation and the rim of the disc in 2 seconds.
one year ago

Answers : (1)

3008 Points

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

Area swept by radius vector during one revolution


= π(10 cm)2

= 100π cm2

= π×10-2 m2

Area swept in one second,

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

   = π×10-2×1000

A = 600 πm2

Rate of change of magnetic flux

= dϕB/dt

= BA

= 0.2×600π

= 120π Wb

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

E = dϕB/dt

Therefore, magnitude of e.m.f. is

E = 120 πV


one year ago
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