SATHEE: Electromagnetic Induction and Alternating Current 1 Question 14

Electromagnetic Induction and Alternating Current 1 Question 14

14. A metal bar $A B$ can slide on two parallel thick metallic rails separated by a distance $l$ . A resistance $R$ and an inductance $L$ are connected to the rails as shown in the figure. A long straight

wire, carrying a constant current $I_{0}$ is placed in the plane of the rails as shown. The bar $A B$ is held at rest at a distance $x_{0}$ from the long wire. At $t = 0$ , it made to slide on the rails away from the wire. Answer the following questions.

$(2002, 5 M)$

(a) Find a relation among $i, \frac{d i}{d t}$ and $\frac{d φ}{d t}$ , where $i$ is the current in the circuit and $φ$ is the flux of the magnetic field due to the long wire through the circuit.

(b) It is observed that at time $t = T$ , the metal bar $A B$ is at a distance of $2 x_{0}$ from the long wire and the resistance $R$ carries a current $i_{1}$ . Obtain an expression for the net charge that has flown through resistance $R$ from $t = 0$ to $t = T$ .

(c) The bar is suddenly stopped at time $T$ . The current through resistance $R$ is found to be $i_{1} / 4$ at time $2 T$ . Find the value of $L / R$ in terms of the other given quantities.

Show Answer

Answer:

Correct Answer: 14. (a) $\frac{d φ}{d t} = i R + L \frac{d i}{d t}$ (b) $\frac{1}{R} \frac{μ_{0} I_{0} l}{2 π} (\ln (2) - L i_{1})$ (c) $\frac{T}{\ln (4)}$

Solution:

(a) Applying Kirchhoff’s second law, we get

$\begin{aligned} \frac{d φ}{d t} - i R - L \frac{d i}{d t} = 0 \\ \frac{d φ}{d t} = i R + L \frac{d i}{d t} \end{aligned}$

This is the desired relation between $i, \frac{d i}{d t}$ and $\frac{d φ}{d t}$ .

(b) Eq. (i) can be written as

$d φ = i R d t + L d i$

Integrating, we get

$\begin{aligned} Δ φ & = R Δ q + L i_{1} \\ Δ q & = \frac{Δ φ}{R} - \frac{L i_{1}}{R} \end{aligned}$

$Here, Δ φ = φ_{f} - φ_{i} = \int_{x = 2 x_{0}}^{x = x_{0}} \frac{μ_{0}}{2 π} \frac{I_{0}}{x} l d x = \frac{μ_{0} I_{0} l}{2 π} \ln$

So, from Eq. (ii) charge flown through the resistance upto time $t = T$ , when current $i_{1}$ , is

$Δ q = \frac{1}{R} \frac{μ_{0} I_{0} l}{2 π} \ln (2) - L i_{1}$

$i = i_{0} e^{- t / τ_{L}}$

Here, $i = \frac{i_{1}}{4}, i_{0} = i_{1}, t = (2 T - T) = T$ and $τ_{L} = \frac{L}{R}$

Substituting these values in Eq. (iii), we get

$τ_{L} = \frac{L}{R} = \frac{T}{\ln 4}$

पिछला

अगला

Electromagnetic Induction and Alternating Current 1 Question 14

14. A metal bar $A B$ can slide on two parallel thick metallic rails separated by a distance $l$ . A resistance $R$ and an inductance $L$ are connected to the rails as shown in the figure. A long straight

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Electromagnetic Induction and Alternating Current 1 Question 14

14. A metal bar AB can slide on two parallel thick metallic rails separated by a distance l. A resistance R and an inductance L are connected to the rails as shown in the figure. A long straight

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Menu

14. A metal bar $A B$ can slide on two parallel thick metallic rails separated by a distance $l$ . A resistance $R$ and an inductance $L$ are connected to the rails as shown in the figure. A long straight