SATHEE: Electromagnetic Induction and Alternating Current 2 Question 10

Electromagnetic Induction and Alternating Current 2 Question 10

11. Two parallel vertical metallic rails $A B$ and $C D$ are separated by $1 m$ . They are connected at two ends by resistances $R_{1}$ and $R_{2}$ as shown in figure. A horizontal metallic bar of mass $0.2 k g$ slides without friction vertically down the rails under the action of gravity.

There is a uniform horizontal magnetic field of $0.6 T$ 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 $0.76 W$ and $1.2 W$ respectively. Find the terminal velocity of the bar $L$ and the values of $R_{1}$ and $R_{2}$ . (1994,6M)

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Answer:

Correct Answer: 11. $v = 1 m / s, R_{1} = 0.47 Ω, R_{2} = 0.3 Ω$

Solution:

Let the magnetic field be perpendicular to the plane of rails and inwards $\otimes$ . If $v$ be the terminal velocity of the rails, then potential difference across $E$ and $F$ would be $B v L$ with $E$ at lower potential and $F$ at higher potential. The equivalent circuit is shown in figure (2). In figure (2)

$\begin{aligned} i_{1} = \frac{e}{R_{1}} \\ i_{2} = \frac{e}{R_{2}} \end{aligned}$

Power dissipated in $R_{1}$ is $0.76 W$

Therefore $e i_{1} = 0.76 W$

Similarly, $e i_{2} = 1.2 W$

Now, the total current in bar $E F$ is

$i = i_{1} + i_{2}$

(from $E$ to $F$ ) …(v)

(1)

(2) Under equilibrium condition, magnetic force $(F_{m})$ on bar $E F$ = weight $(F_{g})$ of bar $E F$

i.e.

$F_{m} = F_{g} or i L B = m g$

From Eq. (vi) $i = \frac{m g}{L B} = \frac{(0.2) (9.8)}{(1.0) (0.6)}$

$i = 3.27 A$

Multiplying Eq. (v) by $e$ , we get

$e i = e i_{1} + e i_{2}$

$\begin{aligned} = (0.76 + 1.2) [From Eqs. (iii) and (iv)] \\ = 1.96 W \\ e = \frac{1.96}{i} V = \frac{1.96}{3.27} \\ or e = 0.6 V \\ But since e = B v L \\ v = \frac{e}{B L} = \frac{(0.6)}{(0.6) (1.0)} = 1.0 m / s \end{aligned}$

Hence, terminal velocity of bar is $1.0 m / s$ .

Power in $R_{1}$ is $0.76 W$

$\begin{aligned} ∴ 0.76 = \frac{e^{2}}{R_{1}} \Rightarrow ∴ R_{1} = \frac{e^{2}}{0.76} = \frac{(0.6)^{2}}{0.76} \\ = 0.47 Ω \Rightarrow R_{1} = 0.47 Ω \\ Similarly, R_{2} = \frac{e^{2}}{1.2} = \frac{(0.6)^{2}}{1.2} = 0.3 Ω \\ R_{2} = 0.3 Ω \end{aligned}$

Electromagnetic Induction and Alternating Current 2 Question 10

11. Two parallel vertical metallic rails $A B$ and $C D$ are separated by $1 m$ . They are connected at two ends by resistances $R_{1}$ and $R_{2}$ as shown in figure. A horizontal metallic bar of mass $0.2 k g$ slides without friction vertically down the rails under the action of gravity.

Answer:

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Electromagnetic Induction and Alternating Current 2 Question 10

11. Two parallel vertical metallic rails AB and CD are separated by 1m. They are connected at two ends by resistances R1 and R2 as shown in figure. A horizontal metallic bar of mass 0.2kg slides without friction vertically down the rails under the action of gravity.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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11. Two parallel vertical metallic rails $A B$ and $C D$ are separated by $1 m$ . They are connected at two ends by resistances $R_{1}$ and $R_{2}$ as shown in figure. A horizontal metallic bar of mass $0.2 k g$ slides without friction vertically down the rails under the action of gravity.