Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
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.
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 m2Area swept in one second,A = (area swept in one revolution) × (Number of revolutions ) = π×10-2×1000A = 600 πm2Rate of change of magnetic flux= dϕB/dt= BA= 0.2×600π= 120π WbAccording to Faraday’s law, magnitude of induced e.m.f is,E = dϕB/dtTherefore, magnitude of e.m.f. isE = 120 πV
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 )
= 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
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -