Question 40: Electromagnetic Induction

Question

$X P Q Y$ is a vertical smooth long loop having a total resistance $R$ where $P X$ is parallel to $Q Y$ and separation between them is $l$. A constant magnetic field $B$ perpendicular to the plane of the loop exists in the entire space. A rod $C D$ of length $L(L>l)$ and mass $m$ is made to slide down from rest under the gravity as shown in figure. The terminal speed acquired by the rod is $\_\_\_\_$ $\mathrm{m} / \mathrm{s} .(\mathrm{g}=$ acceleration due to gravity)

Accepted Answer

As the rod falls under gravity with a downward velocity v , it cuts through the perpendicular magnetic field lines. This changing magnetic flux through the closed loop XPQY induces an EMF according to Faraday's Law of Induction.
Let the distance from the top resistor R to the rod be y.
Even though the rod has length $L$, only the length $l$ between the vertical rails completes the electrical circuit. The overhanging parts do not contribute to the closed loop's area.
So, the area of the rectangular loop formed by the rails, the resistor, and the active part of the rod is $\mathrm{A}=\mathrm{l} 	imes \mathrm{y}$.
The magnetic flux $\Phi$ passing through this area is :
$$ \Phi=\mathrm{B} \cdot \mathrm{~A}=\mathrm{B} \cdot \mathrm{l} \cdot \mathrm{y} $$
According to Faraday's Law, the magnitude of the induced EMF (e) is the rate of change of magnetic flux :
$$ e=\left|\frac{d \Phi}{d t}ight|=\left|\frac{d(B \cdot l \cdot y)}{d t}ight| $$
$$\Rightarrow $$ $$ e=B l\left|\frac{d y}{d t}ight| $$
The rate of change of position $y$ with respect to time is simply the velocity $v$ of the rod:
$$ \mathrm{e}=\mathrm{Blv} $$
The closed loop has a total resistance R. According to Ohm's Law, the induced current I flowing through the circuit is:
$$ I=\frac{e}{R}=\frac{B l v}{R} $$
When a current-carrying conductor is placed in a magnetic field, it experiences a magnetic force. The formula for the force on a straight segment of wire is $\overrightarrow{\mathrm{F}}_{\mathrm{m}}=\mathrm{I}\left(\overrightarrow{\mathrm{l}}_{\mathrm{eff}} 	imes \overrightarrow{\mathrm{B}}ight)$.
The current I only flows through the part of the rod between the rails (length l). The overhanging parts carry no current because they are open circuits.
Therefore, the effective length is l , not L .
Since the length vector is perpendicular to the magnetic field, the magnitude of the force is $\mathrm{F}_{\mathrm{m}}=\mathrm{IlB}$.
$$ \mathrm{F}_{\mathrm{m}}=\left(\frac{\mathrm{Blv}}{\mathrm{R}}ight) \mathrm{lB} $$
$$\Rightarrow $$ $$ F_m=\frac{B^2 l^2 v}{R} $$
According to Lenz's Law, the direction of the induced current will be such that the resulting magnetic force opposes the change that caused it. Since the rod is falling down (increasing the area), the magnetic force will act upwards to oppose the downward motion.
The net downward force is :
$\mathrm{F}_{\mathrm{net}}=\mathrm{mg}-\mathrm{F}_{\mathrm{m}}$
$$\Rightarrow $$ $$ \mathrm{ma}=\mathrm{mg}-\frac{\mathrm{B}^2 \mathrm{l}^2 \mathrm{v}}{\mathrm{R}} $$
When the rod is first released from rest $(\mathrm{v}=0)$, the upward magnetic force is zero, and the rod accelerates downwards at g . As its velocity v increases, the upward magnetic force increases.
Eventually, the upward magnetic force becomes exactly equal to the downward weight. At this point, the net force becomes zero, and the acceleration (a) becomes zero. The rod then continues to fall at a constant, maximum velocity known as the terminal velocity $\left(\mathrm{v}_{\mathrm{T}}ight)$.
At terminal velocity $\left(\mathrm{v}=\mathrm{v}_{\mathrm{T}}ight), \mathrm{a}=0$ :
$$ 0=\mathrm{mg}-\frac{\mathrm{B}^2 \mathrm{l}^2 \mathrm{v}_{\mathrm{T}}}{\mathrm{R}} $$
$$\Rightarrow $$ $\mathrm{mg}=\frac{\mathrm{B}^2 \mathrm{l}^2 \mathrm{v}_{\mathrm{T}}}{\mathrm{R}}$
$$\Rightarrow $$ $$ \mathrm{v}_{\mathrm{T}}=\frac{\mathrm{mgR}}{\mathrm{~B}^2 \mathrm{l}^2} $$
Therefore, the terminal speed of rod is $\frac{m g R}{B^2 1^2}$. Hence, the correct option is (A).

Answer

$\frac{m \mathrm{~g} R}{B^2 l^2}$

Answer

$\frac{2 m g R}{B^2 L^2}$

Answer

$\frac{2 m g R}{B^2 l^2}$

Answer

$\frac{8 m g R}{B^2 l^2}$

Electromagnetic Induction

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