A conductor of length 1 metre falls freely under gravity from height 10 metre .Find the induced EMF across the conductor.when(i)the conductor is held vertically.(ii)when the conductor is held horizontally in east-west direction.Given horizontal component of Earth magnetic field =3×10^-5Wb/m^2.
Answers
Answer:
ANSWER
(a)
By faraday's laws of electromagnetic induction,
V=
dt
dϕ
V=B
dt
dA
V=Blv as dA/dt=lv
This is the induced emf across the ends of the rod.
(b)
We have the force on the free electrons from Q to P as
F
b
=
qv
×
B
where v is the velocity of the rod as well as of the free electrons inside it, B is the uniform magnetic field. The free electrons will move towards P and positive charge will appear at Q. An electrostatic field is developed from Q to P in the wire which exerts a force
F
e
=q
E
on each electron. The charge keeps on gathering until
F
b
=
F
e
and the resultant force on each electron is zero.
∣q
v
×
B
∣=∣q
E
∣
⇒vB=E
The potential difference between Q and P is then
V=El=vBl
which is maintained by the magnetic force of the moving electron producing an emf, e=Bvl
solution