Two parallel vertical metallic rails AB and CD are separated by 1 m. They are connected at two ends

Q: Why does the bar reach a terminal velocity?

As the bar falls, its velocity increases, which increases the induced EMF and the current. This larger current results in a stronger upward magnetic force. Eventually, this upward force perfectly balances the downward gravitational force, resulting in zero net acceleration and a constant terminal velocity.

Q: Why do we model the sliding bar as a battery?

According to Faraday's Law of Induction, a conductor moving through a magnetic field experiences an induced EMF. This EMF drives current through the circuit just like a chemical battery would, with the EMF value given by $e = BvL$.

Q: How do we determine the polarity of the induced EMF?

You can use Fleming's Right-Hand Rule or the Lorentz force equation $\vec{F} = q(\vec{v} \times \vec{B})$. With velocity downwards and magnetic field inwards, the force on positive charges is towards the right, making the right end of the bar the positive terminal.

Q: Why is the total power equal to the sum of the powers in the resistors?

By the principle of conservation of energy, the mechanical power converted into electrical power by the falling bar must be entirely dissipated as heat in the resistors (assuming ideal, frictionless rails).

Q: Why did we use $P = e^2/R$ instead of $P = i^2 R$?

Since the two resistors are connected in parallel across the sliding bar, they both experience the same potential difference, which is the induced EMF $e$. Using $P = e^2/R$ allows us to directly solve for $R$ without needing to calculate the individual branch currents first.

Question

Two parallel vertical metallic rails $AB$ and $CD$ are separated by $1	ext{ m}$. They are connected at two ends by resistances $R_1$ and $R_2$ as shown in figure. A horizontal metallic bar of mass $0.2	ext{ kg}$ slides without friction vertically down the rails under the action of gravity.
There is a uniform horizontal magnetic field of $0.6	ext{ T}$ perpendicular to the plane of the rails. It is observed that when the terminal velocity is attained, the powers dissipated in $R_1$ and $R_2$ are $0.76	ext{ W}$ and $1.2	ext{ W}$ respectively. Find the terminal velocity of the bar $L$ and the values of $R_1$ and $R_2$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a metallic bar sliding down two vertical rails. As it falls through the magnetic field, an electromotive force is induced, driving a current through the circuit. We are given a mass of $m = 0.2 	ext{ kg}$, a length of $L = 1 	ext{ m}$, and a uniform magnetic field of $B = 0.6 	ext{ T}$.

As the bar accelerates downwards due to gravity, its velocity increases. According to Faraday's Law, this increasing velocity generates a larger motional EMF, which in turn drives a larger current through the bar. This current interacts with the external magnetic field to produce an upward magnetic force.

### The Master Equation: Terminal Velocity

The bar eventually reaches a constant terminal velocity. Why? Because the downward pull of gravity is perfectly balanced by the upward magnetic force acting on the current-carrying bar. This is a classic equilibrium condition where the net force becomes zero.

Mathematically, we equate the magnetic force to the gravitational force:

$$ F_m = F_g $$

$$ iLB = mg $$

By rearranging this equation, we can easily calculate the total current flowing through the bar:

$$ i = \frac{mg}{LB} = \frac{0.2 	imes 9.8}{1 	imes 0.6} = \frac{1.96}{0.6} 	ext{ A} $$

### The Equivalent Circuit: A Moving Battery

To understand the power distribution, let's draw an equivalent circuit. The moving bar acts just like a battery with an EMF equal to $e = BvL$. The total current $i$ splits into two branches, flowing through the top resistor $R_1$ and the bottom resistor $R_2$.

We are given the power dissipated in both resistors: $P_1 = 0.76 	ext{ W}$ and $P_2 = 1.2 	ext{ W}$. The total power generated by our 'moving battery' must equal the sum of the power dissipated in these resistors due to the conservation of energy.

$$ P_{	ext{total}} = P_1 + P_2 = 0.76 + 1.2 = 1.96 	ext{ W} $$

### Power Dynamics and Induced EMF

This total power is also equal to the product of the induced EMF and the total current ($P = e \cdot i$). Since we know the total power is $1.96 	ext{ W}$ and the total current is $\frac{1.96}{0.6} 	ext{ A}$, we can divide them to find the induced EMF.

$$ e = \frac{P_{	ext{total}}}{i} = \frac{1.96}{\frac{1.96}{0.6}} = 0.6 	ext{ V} $$

Notice how beautifully the $1.96$ cancels out! The EMF comes out to be exactly $0.6 	ext{ V}$.

Now for the exciting part! We know the EMF is $e = BvL$. Rearranging this for velocity, we substitute our EMF of $0.6 	ext{ V}$. The magnetic field is $0.6 	ext{ T}$ and length is $1 	ext{ m}$.

$$ v = \frac{e}{BL} = \frac{0.6}{0.6 	imes 1} = 1.0 	ext{ m/s} $$

This gives us a terminal velocity of exactly $1.0 	ext{ m/s}$.

### The Final Calculation

Finally, let's find the resistance values. Using the power formula $P = \frac{e^2}{R}$, we can isolate $R$. Plugging in the EMF of $0.6 	ext{ V}$ and the respective power for each resistor, we get:

$$ R_1 = \frac{e^2}{P_1} = \frac{0.6^2}{0.76} = 0.47 \ \Omega $$

$$ R_2 = \frac{e^2}{P_2} = \frac{0.6^2}{1.2} = 0.3 \ \Omega $$

**And there we have it!** The bar falls with a steady terminal velocity of $1.0 	ext{ m/s}$, and the resistances are $0.47 \ \Omega$ and $0.3 \ \Omega$. A beautiful application of electromagnetic induction and mechanics!

Visualized Solution

$Analyzing the Setup$

$Condition for Terminal Velocity$

$Calculating Total Current$

$Equivalent Circuit$

$Total Power Dissipated$

$Calculating Induced EMF$

$Calculating Terminal Velocity$

$Calculating Resistances$

$Final Answer$

The Sigma Insight: Motional EMF

Analyzing the Setup

The Master Equation

The Equivalent Circuit

Power Dynamics and Induced EMF

The Final Calculation

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A thin semicircular conducting ring of radius $R$ is falling with its plane vertical in a horizontal magnetic induction $B$ . At the position $M N Q$ the speed of the ring is $v$ and the potential difference developed across the ring is

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Condition for Terminal Velocity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Total Current

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equivalent Circuit

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Power Dissipated

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Induced EMF

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Terminal Velocity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Resistances

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer