Figure shows a metal rod PQ resting on the smooth rails AB and positioned between the poles of a permanent magnet. The rails,the rod, and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, B = 0.50 T, resistance of the closed loop containing the rod = 9.0 mΩ. Assume the field to be uniform.
(a) Suppose K is open and the rod is moved with a speed of 12cms−1 in the direction shown. Give the polarity and magnitude of the induced emf.
(b) Is there an excess charge built up at the ends of the rods when K is open? What if K is closed?
(c) With K open and the rod moving uniformly, there is no net force on the electrons in the rod PQ even though they do experience magnetic force due to the motion of the rod. Explain.
(d) What is the retarding force on the rod when K is closed?
(e) How much power is required (by an external agent) to keep the rod moving at the same speed (=12cms−1) when K is closed? How much power is required when K is open?
(f) How much power is dissipated as heat in the closed circuit? What is the source of this power?
(g) What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular?
(a) Induce EMF is e=Blv=9×10−3VThe polarity of the induced emf is such that end P shows positive while end Q shows negative ends.
(b)
Yes;
when key K is closed, excess charge is maintained by the continuous
flow of current.
When
key K is open, there is excess charge built up at both ends of the
rods.
When
key K is closed, excess charge is maintained by the continuous flow
of current.
(c)
Magnetic
force is cancelled by the electric force set-up due to the excess
charge of opposite
nature at both ends of the rod.
(d)
Retarding
force exerted on the rod, F=IBl
I=e/R=9×10−3/9×10−3=1A
so,
F=1×0.5×0.15=75×10−3N
(e)
Retarding
force, F=BIl
where,
I=e/R=1A
so,
F=75×10−3N
Also,
Power P=Fv=75×10−3×0.12=9mW
(f)
Power
dissipated = I2R=9mW
The
source of this power external agent.
(g)
Zero.
As, in this case, no emf is induced in the coil because the motion of
the rod does not cut across the field lines.
(b)
Yes; when key K is closed, excess charge is maintained by the continuous flow of current.
When key K is open, there is excess charge built up at both ends of the rods.
When key K is closed, excess charge is maintained by the continuous flow of current.
(c)
Magnetic force is cancelled by the electric force set-up due to the excess charge of opposite nature at both ends of the rod.
(d)
Retarding force exerted on the rod, F=IBl
I=e/R=9×10−3/9×10−3=1A
so, F=1×0.5×0.15=75×10−3N
(e)
Retarding force, F=BIl
where, I=e/R=1A
so, F=75×10−3N
Also, Power P=Fv=75×10−3×0.12=9mW
(f)
Power dissipated = I2R=9mW
The source of this power external agent.
(g)
Zero. As, in this case, no emf is induced in the coil because the motion of the rod does not cut across the field lines.
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