A closed coil having 100 turns is rotated in a uniform magnetic field b=4*104 T about a diameter which is - Physics - Electromagnetic Induction

Question

A closed coil having 100 turns is rotated in a uniform magnetic
field b=4*104 T about a diameter which is perpendicular
to the field The angular velocity of rotation is 300 rev/min the
area of the coil is 25 cm2 and its resistance is 4 ohm
find- 1 average emf developed in half a turn from a position where
the coil is perpendicular to the magnetic field 2 averageemf in
full turn 3 net charge displaced in part 1

Himanshu · Accepted Answer

$(a) A v e r a g e e m f = \frac{2 N B A}{t} = \frac{2 \times 100 \times 4 \times 10^{- 4} \times 25 \times 10^{- 4}}{\frac{1}{600} \times 60} = 2 \times 10^{- 3} V (b) I n f u l l t u r n, t h e n e t c h a n g e o f f l u x w i l l b e z e r o . T h e r e f o r e, a v e r a g e e m f w i l l b e 0.$

$(c) I n d u c e d c h \arg e = \frac{2 N B A}{R} = \frac{2 \times 100 \times 4 \times 10^{- 4} \times 25 \times 10^{- 4}}{4} = 5 \times 10^{- 5} C$