NCERT Class XII Chapter
Electromagnetic Induction
Questions With Solutions
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Question : 14
Total: 17
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 12 cms – 1
(b) Is there an excess charges 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 (= 12 cm s–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) Suppose K is open and the rod is moved with a speed of 12 cm
(b) Is there an excess charges 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 (= 12 cm s–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?
Solution:
Here rails, rod and magnetic field are in three mutually perpendicular directions.
(a) Switch K is open and rod moves with speed of 12 cms – 1
Induced emf/motional emf
ε = Bvl
ε = 0.5 × 12 ×10 – 2 10 – 2
(b) When the K is open, upper end of the rod become positively charge, and lower end become negatively charged.
When the K is closed the charge flows in closed circuit but the excess charge is maintained by the flow of charge in the moving rod under magnetic force.
(c) In the state when K is open very soon a stage is reached when force due to electric field which is due to potential difference induced balances the magnetic force on electrons
eE = Bev or
e
Motional emf V = Bvl.
(d) When the key is closed the current flows in a loop and the current carrying wire experience a retarding force in the magnetic field.
F m
where I =
F m 10 − 2
(e) To keep the rod moving in closed circuit at constant speed the force required is F = 0.075 N.
So, power required
P =
⋅
= Fv cos 0° = Fv or P = 0.075 × 12 ×10 – 2
when key K is open, no current flows and hence no retarding force, so no power is required to move at constant speed.
(f) Power lost in closed circuit due to flow of current
P =I 2 R ( 1 ) 2 10 – 3
Power provided by external force to move the rod at constant speed is the source of this power lost.
(g) If
(a) Switch K is open and rod moves with speed of 12 cm
Induced emf/motional emf
ε = Bvl
ε = 0.5 × 12 ×
(b) When the K is open, upper end of the rod become positively charge, and lower end become negatively charged.
When the K is closed the charge flows in closed circuit but the excess charge is maintained by the flow of charge in the moving rod under magnetic force.
(c) In the state when K is open very soon a stage is reached when force due to electric field which is due to potential difference induced balances the magnetic force on electrons
eE = Bev or
e
Motional emf V = Bvl.
(d) When the key is closed the current flows in a loop and the current carrying wire experience a retarding force in the magnetic field.
where I =
(e) To keep the rod moving in closed circuit at constant speed the force required is F = 0.075 N.
So, power required
P =
= Fv cos 0° = Fv or P = 0.075 × 12 ×
when key K is open, no current flows and hence no retarding force, so no power is required to move at constant speed.
(f) Power lost in closed circuit due to flow of current
P =
Power provided by external force to move the rod at constant speed is the source of this power lost.
(g) If
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