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Standard XII
Physics
Gauss Law for Magnetism
Question

Above figure shows a metal rod $$PQ$$ resting on the smooth rails $$AB$$ and positioned between the poles of a permenent magnet. The rails, the rod, and the magnetic filed are in three mutual perpendicular directions. A galvanometer $$G$$ connects the rail through a switch $$K$$. Length of the rod is $$15cm$$, $$B=0.50T$$, resistanbce of the closed-loop containing the rod is $$9.0$$. Assume the filed to be uniform.
(a) Suppose $$K$$ is open and the rod is moved with a speed of $$12cm$$ $$s^{-1}$$ in the direction shown. Give the polarity and magnitude of the induced emf.
(b) Is there an excess charge build up at the ends of the rods when $$K$$ is open? What is $$K$$ is closed?
(c) With $$K$$ open and the rod moving uniformly, there is no net force on the electrons in the rod $$PQ$$ even through 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 ($$=12cm$$ $$s^{-1}$$) 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?

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Solution
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Here, $$B=0.50T;l=15cm =15*10^{-2}m$$;
$$R=9.0mQ=9.0*10^{3}fl$$
(a) Now, $$e=BVl$$
Here, $$V=12cm$$ $$s^{-1}=12*10^{-2}ms^{-1}$$
$$\therefore e=0.50*12*10^{-2}*15*10^{2}=9*10^{-3}V$$
If $$q$$ is change on as electron, then the elctrons in the rod will experience magnetic Lorentz force $$-q[\underset{v}{\rightarrow}+\underset{B}{\rightarrow}]P.Q$$. Hence, the end $$P$$ of the rod will become positive and the end $$Q$$ will become negative.
(b) When the switch $$K$$ is open, the elctron collect at the end $$Q$$. Therefore, excess change is build up at the end $$Q$$. However, when the switch $$K$$ is closed, the accumulated charge at the end $$Q$$ flows through the circuit.
(c) The magnetic Lorentz force on electron is cancelled by the electronic force acting on it due to the electronic filed set up accross the two ends due to accumulation of positive and negative charges at the end $$P$$ and $$Q$$ respectively.
(d) Retarding force, $$F=BIl=0.50*\frac{e}{R}*15*10^{-2}$$
$$=0.50*\frac{9*10^{-3}}{9*10^{-3}}*15*10^{-2}=7.5*10^{-2}N$$.
(e) When the switch $$K$$ is closed, power required by theb external agent against the retarding force,
$$P=Fv=7.5*10^{-2}*12*10^{-2}=9*10^{-3}W$$.
(f) Power dissipated as heat,
$$\frac{e^{2}}{R}=\frac{[9*10^{-3}]^{2}}{9*10^{-3}}=9*10^{-3}W$$/
The sorce of this power is the power of the external agent against the retarding force,
$$P=Fv=7.5*10^{-2}*12*10^{-2}=9*10^{-3}W$$.
(g) The motion of rod does not cut field lines, hence no induced e.m.f. is produced.

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Q1
Above figure shows a metal rod $$PQ$$ resting on the smooth rails $$AB$$ and positioned between the poles of a permenent magnet. The rails, the rod, and the magnetic filed are in three mutual perpendicular directions. A galvanometer $$G$$ connects the rail through a switch $$K$$. Length of the rod is $$15cm$$, $$B=0.50T$$, resistanbce of the closed-loop containing the rod is $$9.0$$. Assume the filed to be uniform.
(a) Suppose $$K$$ is open and the rod is moved with a speed of $$12cm$$ $$s^{-1}$$ in the direction shown. Give the polarity and magnitude of the induced emf.
(b) Is there an excess charge build up at the ends of the rods when $$K$$ is open? What is $$K$$ is closed?
(c) With $$K$$ open and the rod moving uniformly, there is no net force on the electrons in the rod $$PQ$$ even through 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 ($$=12cm$$ $$s^{-1}$$) 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?

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Q2

Figure 6.20 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 cm s−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 (=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?

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Q3
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