Figure showsa metal rod $\mathbf{P Q}$ resting on the smooth rails $\mathbf{A B}$ 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 $\mathrm{G}$ connects the rails through a switch $\mathrm{K}$. Length of the rod $=15 \mathrm{~cm}, B=0.50 \mathrm{~T}$, resistance of the closed loop containing the rod $=9.0 \mathrm{~m} \Omega$. Assume the field to be uniform.
Suppose $K$ is open and the rod is moved with a speed of $12 \mathrm{~cm} \mathrm{~s}^{-1}$ in the direction shown. Give the polarity and magnitude of the induced emf.
(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 $\mathrm{K}$ open and the rod moving uniformly, there is no net force on the electrons in the rod $\mathrm{PQ}$ 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 $\left(=12 \mathrm{~cm} \mathrm{~s}^{-1}\right)$ 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?

Options:

None

Solution:
Here rails, rod and magnetic field are in three mutually perpendicular directions.
(a) Switch $\mathrm{K}$ is open and rod moves with speed of $12 \mathrm{~cm}$ $\mathrm{s}^{-1}$.
Induced emf/motional emf $e=B v l=0.5 \times 12 \times 10^{-2} \times 15 \times 10^{-2}=\mathrm{mV}$
(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 $\mathrm{K}$ 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.
$$
\begin{gathered}
e E=B e v \\
e \frac{V}{l}=B e v
\end{gathered}
$$
Motional emf $V=B v l$.
(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.
$$
F_m=I B l
$$
where $I=\frac{B v l}{R}=\frac{9 \times 10^{-3}}{9 \times 10^{-3}}=1 \mathrm{~A}$
$$
F_m=1 \times 0.5 \times 15 \times 10^{-2}=0.075 \mathrm{~N}
$$
(e) To keep the rod moving in closed circuit at constant speed the force required is $f=0.005 \mathrm{~N}$.
So, power required $P=\vec{F} \cdot \vec{v}=$ fv $\cos 0^{\circ}=f v$
$$
P=0.075 \times 12 \times 10^{-2}=9 \mathrm{~mW}
$$
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 to current $P=F^2 \mathrm{R}=(1)^2 \times 9 \times 10^{-3}=9 \mathrm{~mW}$
Power provided by external force to move the rod at constant speed is the source of this power lost.
(g) If $\vec{B}$ is parallel to rails, the induced/motional emf will be zero.

Topic: Physics, Electromagnetic Induction