A CIRCULAR COIL OF 20 TURNS OF RADIUS 10.0 cm IS PERPENDICULAR TO A 0.30 tesla MAGNETIC FIELD. HOW MUCH INDUCED emf IS CREATED WHEN THE COIL IS ROTATED IN 0.12 s, SO ITS FACE IS PARALLEL TO THE FIELD?

1.57 V

Explanation:

The induced emf ε = -ΔΦ/Δt where Φ = magnetic flux = NABcosθ where N = number of turns of circular coil = 20, A = area of coil = πr² where r = radius of coil = 10.0 cm = 0.1 m , B = strength  of magnetic field = 0.30 T,θ = angle between normal to are and magnetic field and Δt = change in time = 0.12 s

Now ΔΦ = change in magnetic flux through circular coil = Φ' - Φ

Φ = magnetic field when A is perpendicular to B , that is θ = 90°. So, Φ = NABcos90 = 0 and Φ' = magnetic field when A is parallel to B , that is θ = 0°. So, Φ' = NABcos0 = NAB

So, ΔФ = Ф' - Ф

= 0 - NAB

= -NAB

= -Nπr²B

= -20 × π(0.1 m)² × 0.30 T

= -0.19 Wb

So, ε = -ΔΦ/Δt

= -(-0.19 Wb)/0.12 s

= 1.57 Wb/s

= 1.57 V

distance = 150m

displ = root (50 ^2 + 100^2)

root(2500+10000)

root 125000

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j