Question

A) 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 $=15\text{cm},B=0.50T$, resistance of the closed loop containing the rod $=9.0m\Omega$. Assume the field to be uniform.


B) 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 $=15\text{cm},B=0.50T$, resistance of the closed loop containing the rod $=9.0m\Omega$. Assume the field to be uniform.


C) 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 $=15\text{cm},B=0.50T$, resistance of the closed loop containing the rod $=9.0m\Omega$. Assume the field to be uniform.


D) 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 $=15\text{cm},B=0.50T$, resistance of the closed loop containing the rod $=9.0m\Omega$. Assume the field to be uniform.


E) 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 $=15\text{cm},B=0.50T$, resistance of the closed loop containing the rod $=9.0m\Omega$. Assume the field to be uniform.


F) 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 $=15\text{cm},B=0.50T$, resistance of the closed loop containing the rod $=9.0m\Omega$. Assume the field to be uniform.


G) 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 $=15\text{cm},B=0.50T$, resistance of the closed loop containing the rod $=9.0m\Omega$. Assume the field to be uniform.

Answer 1

A) Step 1: Find the magnitude of induced EMF
Given, length of the rod, $l=15\text{cm}=0.15m$
Magnetic field strength, $B=0.50T$
Speed of the rod, $v=12\text{cm/s}=0.12m/s$
Resistance, $R=9.0m\Omega$
Induced emf is given by,
$ϵ=Blv$
$ϵ=0.5\times 0.15\times 0.12$
$ϵ=9\times {10}^{-3}=9mV$

Step 2: Find polarity of induced EMF
If q is magnitude of charge on an electron, then the electrons in the rod will experience magnetic Lorentz force $-q[v\times B]$. Hence, the end P of the rod will become positive, and the end Q will become negative.
Final answer: $9\text{mV},P$ positive end and Q is negative end.


B) Step 1: When K is open When the switch K is open, the electrons collect at the end Q.
Therefore, excess charge is built up at the end Q.

Step 2: When K is closed When the 𝐾 is closed the charge flows in the closed circuit, but the excess charge is maintained by the flow of charge in the moving rod under magnetic force.

Final answer: Yes. When K is closed, the excess charge is maintained by the continuous flow of current.

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 balance the magnetic force on electrons.

$eE=Bev$
$\frac{eV}{l}=Bev$
Motional emf $V=Bvl$

Final answer: Magnetic force is cancelled by the electric force set-up due to the excess charge of opposite signs at the ends of the rod.

D) Step 1: Find the magnitude of induced EMF
Given, length of the rod, $l=15\text{cm}=0.15m$
Magnetic field strength, $B=0.50T$
Speed of the rod, $v=12\text{cm/s}=0.12m/s$
Resistance, $R=9.0m\Omega$
Induced emf is given by,
$ϵ=Blv$
$ϵ=0.5\times 0.15\times 0.12$
$ϵ=9\times {10}^{-3}=9\text{mV}$

Step 2: Find current flowing through the rod.
Given, length of the rod, $l=15\text{cm}=0.15m$
Magnetic field strength, $B=0.50T$
Resistance of the closed loop,
$R=9m\Omega =9\times {10}^{-3}\Omega$
The current flowing through the rod, $I=\frac{ϵ}{R}$
$I=\frac{9\times {10}^{-3}}{9\times {10}^{-3}}$
$I=1A$

Step 3: Calculate retarding force exerted on the rod.
Given, length of the rod, $l=15\text{cm}=0.15m$
Magnetic field strength, $B=0.50T$
The retarding force exerted on the rod,
$F=IlB$
$=1\times 0.15\times 0.5$
$=0.075N=75\times {10}^{-3}N$
Final answer : $75\times {10}^{-3}N$


E) Step 1: Find the magnitude of induced EFM
Given, length of the rod, $l=15\text{cm}=0.15m$
Magnetic field strength, $B=0.50T$
Speed of the rod, $v=12\text{cm/s}=0.12m/s$
Resistance, $R=9.0m\Omega$
Induced emf is given by,
$ϵ=Blv$
$ϵ=0.5\times 0.15\times 0.12$
$ϵ=9\times {10}^{-3}=9\text{mV}$

Step 2: Find current flowing through the rod.
Given, length of the rod, $l=15\text{cm}=0.15m$
Magnetic field strength, $B=0.50T$
Resistance of the closed loop,
$R=9m\Omega =9\times {10}^{-3}\Omega$
The current through the rod, $I=\frac{ϵ}{R}$
$I=\frac{9\times {10}^{-3}}{9\times {10}^{-3}}$
$I=1A$

Step 3: Calculate retarding force exerted on the rod.
Given, length of the rod, $l=15\text{cm}=0.15m$
Magnetic field strength, $B=0.50T$
The retarding force exerted on the rod,
$F=IlB$
$=1\times 0.15\times 0.5$
$=0.075N$
$=75\times {10}^{-3}N$

Step 4: Calculate power is required when K is closed
Given, Speed of the rod,
$v=12\text{cm/s}=0.12m/s$
So, power required, $P=\overrightarrow{F}.\overrightarrow{v}$
$P=Fv\cos 0^{\circ }=Fv$
$P=0.075\times 12\times {10}^{-2}$
$=9mW$

Step 5: Calculate power is required when K is open
When key is open, no current flows and hence no retarding force, so no power is required to move at constant speed.
Final answer : $9mW,0$


F) Step 1: Find the magnitude of induced EMF
Given, length of the rod, $l=15\text{cm}=0.15m$
Magnetic field strength, $B=0.50T$
Speed of the rod, $v=12\text{cm/s}=0.12m/s$
Resistance, $R=9.0m\Omega$
Induced emf is given by,
$ϵ=Blv$
$ϵ=0.5\times 0.15\times 0.12$
$ϵ=9\times {10}^{-3}=9mV$

Step 2: Find current flowing through the rod.
Given, length of the rod, $l=15\text{cm}=0.15m$
Magnetic field strength, $B=0.50T$
Resistance of the closed loop,
$R=9m\Omega =9\times {10}^{-3}\Omega$
The current flowing through the rod, $I=\frac{ϵ}{R}$
$I=\frac{9\times {10}^{-3}}{9\times {10}^{-3}}$
$I=1A$

Step 3: Find the power is dissipated as heat in the closed circuit
Resistance of the closed loop, $R=9m\Omega =9\times {10}^{-3}\Omega$
When the key 𝐾 is closed,
Power dissipated as heat, $P=I^{2}R$
$P=(1{)}^2\times 9\times {10}^{-3}$
$P=9mW$
The source of this power is power provided by an external agent.
Final answer: $9mW$, the source of this power is power provided by an external agent.


G) The induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular is zero, because the motion of the rod doesn’t cross the lines.

Final answer : Zero