5 Marks Questions

Question 515 Marks

An inductor of inductance 2.00H is joined in series with a resistor of resistance $200\Omega$ and a battery of emf 2.00V. At t = 10ms, find

The current in the circuit.
The power delivered by the battery.
The power dissipated in heating the resistor.
The rate at which energy is being stored in magnetic field.

Answer

$\text{L}=2\text{H}, \ \text{R}=200\Omega, \ \text{E}=2\text{V}, \ \text{t}=10\text{ms}$

$\text{l}=\text{l}_0\Big(1-\text{e}^\frac{-\text{t}}{\tau}\Big)$

$=\frac{2}{200}\Big(1-\text{e}^{-10\times10^{-3}\times\frac{200}{2}}\Big)$

$=0.01\Big(1-\text{e}^{-1}\Big)=0.01\Big(1-0.3678\Big)$

$=0.01\times0.632=6.3\text{A}$

Power delivered by the battery.

$=\text{Vl}$

$=\text{El}_0\Big(1-\text{e}^\frac{-\text{t}}{\tau}\Big)=\frac{\text{E}^2}{\text{R}}\Big(1-\text{e}^\frac{-\text{t}}{\tau}\Big)$

$=\frac{2\times2}{200}\Big(1-\text{e}^{-10\times10^{-3}\times\frac{200}{2}}\Big)$

$=0.02\Big(1-\text{e}^{-1}\Big)=0.1264=12\text{mw}.$

Power dissepited in heating the resistor $=\text{l}^2\text{R}$

$=\Big[\text{i}_0\Big(1-\text{e}^\frac{-\text{t}}{\tau}\Big)\Big]^2\text{R}$

$=\big(6.3\text{mA}\big)^2\times200=6.3\times6.3\times200\times10^{-6}$

$=79.38\times10^{-4}=7.938\times10^{-3}=8\text{mA}.$

Rate at which energy is stored in the magnetic field.

$\frac{\text{d}}{\text{dt}}\Big(\frac{1}{2}\text{Ll}^2\Big)$

$=\frac{\text{Ll}^2_0}{\tau}\Big(\text{e}^\frac{-\text{t}}{\tau}-\text{e}^\frac{-2\text{t}}{\tau}\Big)$

$=\frac{2\times10^{-4}}{10^{-2}}\Big(\text{e}^{-1}-\text{e}^{-2}\Big)$

$2\times10^{-2}\Big(0.2325\Big)=0.465\times10^{-2}$

$=4.6\times10^{-3}=4.6\text{mW}.$

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Question 525 Marks

A rectangular frame of wire abcd has dimensions $32cm × 8.0cm$ and a total resistance of $2.0\Omega.$ It is pulled out of a magnetic field $B = 0.020T$ by applying a force of $3.2 \times 10^{-5}N$ (figure). It is found that the frame moves with constant speed. Find

This constant speed.
The emf induced in the loop.
The potential difference between the points a and b.
The potential difference between the points c and d.

Answer

$\text{R}=2.0\Omega, \ \text{B}=0.020\text{T}, \ \text{I}=32\text{cm}=0.32\text{m}$
$\text{B}=8\text{cm}=0.08\text{m}$

$\text{F}=\text{ilB}=3.2\times10^{-5}\text{N}$

$=\frac{\text{B}^2\text{l}^2\text{v}}{\text{R}}=3.2\times10^5$
$\Rightarrow\frac{(0.020)^2\times(0.08)^2\times\text{v}}{2}=3.2\times10^{-5}$
$\Rightarrow\text{v}=\frac{3.2\times10^{-5}\times2}{6.4\times10^{-3}\times4\times10^{-4}}=25\text{m}/\text{s}$

Emf $\text{E}=\text{vBl}=25\times0.02\times0.08=4\times10^{-2}\text{V}$
Resistance per unit length $=\frac{2}{0.8}$

Resistance of part ad/ cb $=\frac{2\times0.72}{0.8}=1.8\Omega$
$\text{V}_{\text{ab}}=\text{iR}=\frac{\text{Blv}}{2}\times1.8=\frac{0.02\times0.08\times25\times1.8}{2}$
$=0.036\text{V}=3.6\times10^{-2}\text{V}$

Resistance of cd $=\frac{2\times0.08}{0.8}=0.2\Omega$

$\text{V}=\text{iR}=\frac{0.02\times0.08\times25\times0.2}{2}=4\times10^{-3}\text{V}$

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Question 535 Marks

Consider a situation similar to that of the previous problem except that the ends of the rod slide on a pair of thick metallic rails laid parallel to the wire. At one end the rails are connected by resistor of resistance R.

What force is needed to keep the rod sliding at a constant speed v?
In this situation what is the current in the resistance R?
Find the rate of heat developed in the resistor.
Find the power delivered by external agent exerting the force on the rod.

Answer

Emf produced due to the current carrying wire $=\frac{\mu_0\text{vi}}{2\pi}\text{In}\Big(\frac{2\text{x}+\text{l}}{2\text{x}-\text{l}}\Big)$

Let current produced in the rod $=\text{i}'=\frac{\mu_0\text{iv}}{2\pi\text{R}}\text{In}\Big(\frac{2\text{x}+\text{l}}{2\text{x}-\text{l}}\Big)$

Force on the wire considering a small portion dx at a distance x

$\text{dF}=\text{i}'\text{B}\text{l} $

$\Rightarrow\text{dF}=\frac{\mu_0\text{iv}}{2\pi\text{r}}\text{In}\Big(\frac{2\text{x}+\text{l}}{2\text{x}-\text{l}}\Big)\times\frac{\mu_0\text{i}}{2\pi\text{x}}\times\text{dx}$

$\Rightarrow\text{dF}=\Big(\frac{\mu_0\text{i}}{2\pi}\Big)^2\frac{\text{V}}{\text{R}}\text{In}\Big(\frac{2\text{x}+\text{l}}{2\text{x}-\text{l}}\Big)\frac{\text{dx}}{\text{x}}$

$\Rightarrow\text{F}=\Big(\frac{\mu_0\text{i}}{2\pi}\Big)^2\frac{\text{V}}{\text{R}}\text{In}\Big(\frac{2\text{x}+\text{l}}{2\text{x}-\text{l}}\Big)\int\limits^{\text{x}+\frac{\text{t}}{2}}_{\text{x}-\frac{\text{t}}{2}}\frac{\text{dx}}{\text{x}}$

$=\Big(\frac{\mu_0\text{i}}{2\pi}\Big)^2\frac{\text{V}}{\text{R}}\text{In}\Big(\frac{2\text{x}+\text{l}}{2\text{x}-\text{l}}\Big)\text{In}\Big(\frac{2\text{x}+\text{l}}{2\text{x}-\text{l}}\Big)$

$=\frac{\text{V}}{\text{R}}\Big[\frac{\mu_0\text{i}}{2\pi}\text{In}\Big(\frac{2\text{x}+\text{l}}{2\text{x}-\text{l}}\Big)\Big]^2$

Current $=\frac{\mu_0\text{In}}{2\pi\text{R}}\text{In}\Big(\frac{2\text{x}+\text{l}}{2\text{x}-\text{l}}\Big)$
Rate of heat developed $=\text{i}^2\text{R}$

$=\Big[\frac{\mu_0\text{iv}}{2\pi\text{R}}\Big(\frac{2\text{x}+\text{l}}{2\text{x}-\text{l}}\Big)\Big]^2\text{R}=\frac{1}{\text{R}}\Big[\frac{\mu_0\text{iv}}{2\pi}\text{In}\Big(\frac{2\text{x}+\text{l}}{2\text{x}-\text{l}}\Big)^2\Big]$

Power developed in rate of heat developed $=\text{i}^2\text{R}$

$=\frac{1}{\text{R}}\Big[\frac{\mu_0\text{vi}}{2\pi}\text{In}\Big(\frac{2\text{x}+\text{l}}{2\text{x}-\text{l}}\Big)\Big]^2$

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Question 545 Marks

Two coils A and B have inductances 1.0H and 2.0H respectively. The resistance of each coil is $10\Omega.$ Each coil is connected to an ideal battery of emf $2.0V$ at $t = 0$ Let $i_A$ and $i_B$ be the currents in the two circuit at time t. Find the ratio $\frac{\text{i}_\text{A}}{\text{i}_\text{B}}$

t = 100ms
t = 200ms
t = 1s.

Answer

$\text{L}_\text{a}=1.0\text{ H }; \text{ L}_\text{B}=2.0\text{H}; \ \text{R}=10\omega$

$\text{t}=0.1\text{s}, \ \tau_\text{A}=0.1, \ \tau_\text{B}=\frac{\text{L}}{\text{R}}=0.2$

$\text{i}_{\text{A}}=\text{i}_0\Big(1-\text{e}^\frac{-\text{t}}{\tau}\Big)$
$=\frac{2}{10}\Big(1-\text{e}^\frac{-0.1\times10}{1}\Big)=0.2\Big(1-\text{e}^{-1}\Big)=0.126424111$
$\text{i}_\text{B}=\text{i}_0\Big(1-\text{e}\frac{-\text{t}}{\tau}\Big)$
$=\frac{2}{10}\Big(1-\text{e}^\frac{-0.1\times10}{2}\Big)=0.2\Big(1-\text{e}^\frac{-1}{2}\Big)=0.078693$
$\frac{\text{i}_\text{A}}{\text{i}_\text{B}}=\frac{0.12642411}{0.78693}=1.6$

$\text{t}=200\text{ms}=0.2\text{s}$

$\text{i}_\text{A}=\text{i}_0\Big(1-\text{e}^\frac{-\text{t}}{\tau}\Big)$
$=0.2\Big(1-\text{e}^\frac{-0.2\times10}{1}\Big)=0.2\times0.864664716=0.172932943$
$\text{i}_\text{B}=0.2\Big(1-\text{e}^\frac{-0.2\times10}{2}\Big)=0.2\times0.632120=0.126424111$
$\frac{\text{i}_\text{a}}{\text{i}_\text{B}}=\frac{0.172932943}{0.126424111}=1.36=1.4$

$\text{t}=1\text{s}$

$\text{i}_\text{A}=0.2\Big(1-\text{e}^\frac{-1\times10}{1}\Big)=0.2\times0.9999546=0.19999092$
$\text{i}_\text{B}=0.2\Big(1-\text{e}^\frac{-1\times10}{2}\Big)=0.2\times0.99326=0.19865241$
$\frac{\text{i}_\text{A}}{\text{i}_\text{B}}=\frac{0.19999092}{19865241}=1.0$

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Question 555 Marks

The rectangular wire-frame, shown in figure, has a width d, mass m, resistance R and a large length. A uniform magnetic field B exists to the left of the frame. A constant force F starts pushing the frame into the magnetic field at $t = 0$.

Find the acceleration of the frame when its speed has increased to v.
Show that after some time the frame will move with a constant velocity till the whole frame enters into the magnetic field. Find this velocity $v_0$.
Show that the velocity at time t is given by $\text{v}=\text{v}_0\Big(1-\text{e}^{-\frac{\text{ft}}{\text{mv}_0}}\Big)$

Answer

emf developed = Bdv (when it attains a speed v)

Current $=\frac{\text{Bdv}}{\text{R}}$
Force $=\frac{\text{B}\text{d}^2\text{v}^2}{\text{R}}$
This force opposes the given force
Net F $=\text{F}-\frac{\text{B}\text{d}^2\text{v}^2}{\text{R}}=\text{RF}-\frac{\text{B}\text{d}^2\text{v}^2}{\text{R}}$
Net acceleration $=\frac{\text{RF}-\text{B}^2\text{d}^2\text{v}}{\text{mR}}$

Velocity becomes constant when acceleration is 0.

$\frac{\text{F}}{\text{m}}-\frac{\text{B}^2\text{d}^2\text{v}_0}{\text{mR}}=0$
$\Rightarrow\frac{\text{F}}{\text{m}}=\frac{\text{B}^2\text{d}^2\text{v}_0}{\text{mR}}$
$\Rightarrow\text{v}_0=\frac{\text{FR}}{\text{B}^2\text{d}^2}$

Velocity at line t

$\text{a}=-\frac{\text{dv}}{\text{dt}}$
$\Rightarrow\int_{0}^{\text{v}}\frac{\text{dv}}{\text{RF}-\text{l}^2\text{B}^2\text{v}}=\int_{0}^{\text{t}}\frac{\text{dt}}{\text{mR}}$
$\Rightarrow\Big[\text{l}_\text{n}\big[\text{RF}-\text{l}^2\text{B}^2\text{v}\big]\frac{1}{-\text{l}^2\text{B}^2}\Big]_0^{\text{v}} \ \Big[\frac{\text{t}}{\text{Rm}}\Big]_0^{\text{t}} $
$\Rightarrow\Big[\text{l}_\text{n}\big(\text{RF}-\text{l}^2\text{B}^2\text{v}\big)\Big]_{0}^{\text{v}}=\frac{-\text{tl}^2\text{B}^2}{\text{Rm}}$
$\Rightarrow\text{l}_\text{n}\big(\text{RF}-\text{l}^2\text{B}^2\text{v}\big)-\text{ln}(\text{RF})=\frac{-\text{t}^2\text{B}^2\text{t}}{\text{Rm}}$
$\Rightarrow1-\frac{\text{l}^2\text{B}^2\text{v}}{\text{Rf}}=\text{e}^{\frac {-\text{l}^2\text{B}^2\text{t}}{\text{Rm}}}$
$\Rightarrow\frac{\text{l}^2\text{B}^2\text{v}}{\text{Rf}}=1-\text{e}^{\frac {-\text{l}^2\text{B}^2\text{t}}{\text{Rm}}}$
$\Rightarrow\text{v}=\frac{\text{FR}}{\text{l}^2\text{B}^2}\Big(1-\text{e}^{\frac{-\text{l}^2\text{B}^2\text{v}_0\text{t}}{\text{Rv}_0\text{m}}}\Big)=\text{v}_0(1-\text{e}^{-\text{Fv}_0\text{m}})$

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Question 565 Marks

Figure shows a square frame of wire having a total resistance r placed coplanarly with a long, straight wire. The wire carries a current i given by $\text{i}=\text{i}_0\sin\omega\text{t}.$ Find

The flux of the magnetic field through the square frame.
The emf induced in the frame.
The heat developed in the frame in the time interval 0 to $\frac{20\pi}{\omega}.$

Answer

Considering an element dx at a dist x from the wire. We have

$\phi=\text{B.A.}$

$\text{d}\phi=\frac{\mu_0\text{i}\times\text{adx}}{2\pi\text{x}}$
$\phi=\int\limits^\text{a}_0\text{d}\phi=\frac{\mu_0\text{ia}}{2\pi}\int\limits^{\text{a}+\text{b}}_\text{b}\frac{\text{dx}}{\text{x}}=\frac{\mu_0\text{ia}}{2\pi}\text{In}\left\{1+\frac{\text{a}}{\text{b}}\right\}$

$\text{e}=\frac{\text{d}\phi}{\text{dt}}=\frac{\text{d}}{\text{dt}}\frac{\mu_0\text{ia}}{2\pi}\text{In}\Big[1+\frac{\text{a}}{\text{b}}\Big]$

$=\frac{\mu_0\text{a}}{2\pi}\text{In}\Big[1+\frac{\text{a}}{\text{n}}\Big]\frac{\text{d}}{\text{dt}}(\text{i}_0\sin\omega\text{t})$
$=\frac{\mu_0\text{ai}_0\omega\cos\omega\text{t}}{2\pi}\text{In}\Big[1+\frac{\text{a}}{\text{b}}\Big]$

$\text{i}=\frac{\text{e}}{\text{r}}=\frac{\mu_0\text{ai}_0\omega\cos\omega\text{t}}{2\pi\text{r}}\text{In}\Big[1+\frac{\text{a}}{\text{b}}\Big]$

$\text{H}=\text{i}^2\text{rt}$
$=\Big[\frac{\mu\text{ai}_0\omega\cos\omega\text{t}}{2\pi\text{r}}\text{In}\Big(1+\frac{\text{a}}{\text{b}}\Big)\Big]^2\times\text{r}\times\text{t}$
$=\frac{\mu_0^2\times\text{a}^2\times\text{i}^2_0\times\omega^2}{4\pi\times\text{r}^2}\text{In}^2\Big[1+\frac{\text{a}}{\text{b}}\Big]\times\text{r}\times\text{}\frac{20\pi}{\omega}$
$=\frac{5\mu_0^2\text{a}^2\text{i}^2_0\times\omega}{2\pi\text{r}}\text{In}^2\Big[1+\frac{\text{a}}{\text{b}}\Big] \ \Big[\therefore\text{t}=\frac{20\pi}{\omega}\Big]$

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Question 575 Marks

A rectangular metallic loop of length l and width b is placed coplanarly with a long wire carrying a current i. The loop is moved perpendicular to the wire with a speed v in the plane containing the wire and the loop. Calculate the emf induced in the loop when the rear end of the loop is at a distance a from the wire. Solve by using Faraday's law for the flux through the loop and also by replacing different segments with equivalent batteries.

Answer

Using Faraday’' law Consider a unit length dx at a distance x$\text{B}=\frac{\mu_0\text{i}}{2\pi\text{x}}$
Area of strip $=\text{b} \ \text{dx}$$\text{d}\phi=\frac{\mu_0\text{i}}{2\pi\text{x}}\text{dx}$
$\Rightarrow\phi=\int\limits^{\text{a}+1}_\text{a}\frac{\mu_0\text{i}}{2\pi\text{x}}\text{bdx}$
$=\frac{\mu_o\text{i}}{2\pi}\text{b}\int\limits^{\text{a}+1}_\text{a}\Big(\frac{\text{dx}}{\text{x}}\Big)=\frac{\mu_0\text{ib}}{2\pi}\log\Big(\frac{\text{a}+\text{l}}{\text{a}}\Big)$
$\text{Emf}=\frac{\text{d}\phi}{\text{dt}}=\text{dt}\Big[\frac{\mu_0\text{ib}}{2\pi}\text{log}\Big(\frac{\text{a}+\text{l}}{\text{a}}\Big)\Big]$
$=\frac{\mu_0\text{ib}}{2\pi}\frac{\text{a}}{\text{a}+\text{l}}\Big(\frac{\text{va}-(\text{a}+\text{l})\text{v}}{\text{a}^2}\Big)$ $\Big($ Where $\frac{\text{da}}{\text{dt}}=\text{V}\Big)$
$=\frac{\mu_0\text{ib}}{2\pi}\frac{\text{a}}{\text{a}+\text{l}}\frac{\text{vl}}{\text{a}^2}=\frac{\mu_0\text{ibvl}}{2\pi(\text{a}+\text{l})\text{a}}$
The velocity of AB and CD creates the emf. since the emf due to AD and BC are equal and opposite to each other.

$\text{B}_{\text{AB}}=\frac{\mu_o\text{i}}{2\pi\text{a}} \ \Rightarrow \ \text{E.m.f.} \ \text{AB}=\frac{\mu_0\text{i}}{2\pi\text{a}}\text{bv}$
Length b, velocity v.$\text{B}_{\text{CD}}=\frac{\mu_0\text{i}}{2\pi(\text{a}+\text{l})}$
$\Rightarrow \text{E.m.f.} \ \text{CD}=\frac{\mu_0\text{ibv}}{2\pi(\text{a}+\text{l})}$
Length b, velocity v.Net emf $=\frac{\mu_0\text{i}}{2\pi\text{a}}\text{bv}-\frac{\mu_0\text{ibv}}{2\pi\text{a}(\text{a}+\text{l})}=\frac{\mu_0\text{ibvl}}{2\pi\text{a}(\text{a}+\text{l})}$

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Question 585 Marks

Consider the situation shown in figure. The wire PQ has mass m, resistance r and can slide on the smooth, horizontal parallel rails separated by a distance l. The resistance of the rails is negligible. A uniform magnetic field B exists in the rectangular region and a resistance R connects the rails outside the field region. At $t = 0$, the wire PQ is pushed towards right with a speed $v_0$. Find

The current in the loop at an instant when the speed of the wire PQ is v.
The acceleration of the wire at this instant.
The velocity vas a functions of x.
The maximum distance the wire will move.

Answer

When the speed is V

Emf = Blv
Resistance = r + r
Current $=\frac{\text{Blv}}{\text{r}+\text{R}}$

Force acting on the wire =ilB

$=\frac{\text{BlvlB}}{\text{r}+\text{R}}=\frac{\text{B}^2\text{l}^2\text{v}}{\text{r}+\text{R}}$
Acceleration on the wire $=\frac{\text{B}^2\text{l}^2\text{v}}{\text{m}(\text{r}+\text{R})}$

$\text{v}=\text{v}_0+\text{at}=\text{v}_0 -\frac{\text{B}^2\text{l}^2\text{v}}{\text{m}(\text{r}+\text{R})}\text{t}$ [force is opposite to velocity]

$=\text{v}_0 -\frac{\text{B}^2\text{l}^2\text{x}}{\text{m}(\text{r}+\text{R})}$

$\text{a}=\text{v}\frac{\text{dv}}{\text{dx}}=\frac{\text{B}^2\text{l}^2\text{x}}{\text{m}(\text{r}+\text{R})}$

$\Rightarrow\text{dx}=\frac{\text{dvm}(\text{R}+\text{r})}{\text{B}^2\text{l}^2}$
$\Rightarrow\text{x}=\frac{\text{m}(\text{R}+\text{r})\text{v}_0}{\text{B}^2\text{l}^2}$

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Question 595 Marks

A conducting wire ab of length l, resistance r and mass m starts sliding at $t = 0$ down a smooth, vertical, thick pair of connected rails as shown in figure. A uniform magnetic field B exists in the space in a direction perpendicular to the plane of the rails.

Write the induced emf in the loop at an instant t when the speed of the wire is v.
What would be the magnitude and direction of the induced current in the wire?
Find the downward acceleration of the wire at this instant.
After sufficient time, the wire starts moving with a constant velocity. Find this velocity $v_m$.
Find the velocity of the wire as a function of time.
Find the displacement of the wire as a function of time.
Show that the rate of heat developed in the wire is equal to the rate at which the gravitational potential energy is decreased after steady state is reached.

Answer

When the speed of wire is V

emf developed = BlV

Induced current is the wire $=\frac{\text{Blv}}{\text{R}}$ (from b to a)
Down ward acceleration of the wire

$=\frac{\text{mg}-\text{F}}{\text{m}}$ due to the current
$=\text{mg}-\text{il}\frac{\text{B}}{\text{m}}=\text{g}-\frac{\text{B}^2\text{l}^2\text{V}}{\text{Rm}}$

Let the wire start moving with constant velocity. Then acceleration = 0

$\frac{\text{B}^2\text{l}^2\text{V}}{\text{Rm}}\text{m}=\text{g}$
$\Rightarrow\text{V}_\text{m}=\frac{\text{gRm}}{\text{B}^2\text{l}^2}$

$\frac{\text{dV}}{\text{dt}}=\text{a}$

$\Rightarrow\frac{\text{dV}}{\text{dt}}=\frac{\text{mg}-\text{B}^2\text{l}^2\frac{\text{v}}{\text{R}}}{\text{m}}$
$\Rightarrow\frac{\text{dv}}{\frac{\text{mg}-\frac{\text{B}^2\text{l}^2\frac{\text{v}}{\text{R}}}{\text{R}}}{\text{m}}}=\text{dt}$
$\Rightarrow\int\limits_{0}^{\text{v}}\frac{\text{mdv}}{\text{mg}-\frac{\text{B}^2\text{l}^2\text{v}}{\text{R}}}=\int\limits_{0}^{\text{t}}\text{dt}$
$\Rightarrow\frac{\text{m}}{\frac{-\text{B}^2\text{l}^2}{\text{R}}}\Big(\log\text{mg}-\frac{\text{B}^2\text{l}^2\text{v}}{\text{R}}\Big)_{0}^{\text{v}}=\text{t}$
$\Rightarrow\frac{-\text{mR}}{\text{B}^2\text{l}^2}=\log\Big[\log\Big(\text{mg}-\frac{\text{B}^2\text{l}^2\text{v}}{\text{R}}\Big)-\log(\text{mg})\Big]=\text{t}$
$\Rightarrow\log\Bigg[\frac{\text{mg}-\frac{\text{B}^2\text{l}^2\text{v}}{\text{R}}}{\text{mg}}\Bigg]=\frac{-\text{tB}^2\text{l}^2}{\text{mR}}$
$\Rightarrow\log\Big[1-\frac{\text{B}^2\text{l}^2\text{v}}{\text{Rmg}}\Big]=\frac{-\text{tB}^2\text{l}^2}{\text{mR}}$
$\Rightarrow1-\frac{\text{B}^2\text{l}^2\text{v}}{\text{Rmg}}=\text{e}^{\frac{-\text{t}\text{B}^2\text{l}^2}{\text{mR}}}$
$\Rightarrow\bigg(1-\text{e}^{\frac{-\text{B}^2\text{l}^2}{\text{mR}}}\bigg)=\frac{\text{B}^2\text{l}^2\text{v}}{\text{Rmg}}$
$\Rightarrow\text{v}=\frac{\text{Rmg}}{\text{B}^2\text{l}^2}\Big(1-\text{e}^{-\frac{\text{B}^2\text{l}^2}{\text{mR}}}\Big)$
$\Rightarrow\text{v}=\text{v}_{\text{m}}\Big(1-\text{e}^{\frac{-\text{gt}}{\text{vm}}}\Big) \ \Big[\text{v}_\text{m}=\frac{\text{Rmg}}{\text{B}^2\text{l}^2}\Big]$

$\frac{\text{ds}}{\text{dt}}=\text{v}\Rightarrow\text{ds}=\text{v} \ \text{dt}$
$\Rightarrow\text{s}=\text{vm}\int\limits_0^\text{t}\Big(1-\text{e}^{\frac{-\text{gt}}{\text{vm}}}\Big)\text{dt}$
$=\text{V}_\text{m}\Big(\text{t}-\frac{\text{V}_\text{m}}{\text{g}}\text{e}^\frac{\text{-gt}}{\text{vm}}\Big)=\Big(\text{V}_\text{m}\text{t}+\frac{\text{V}^2_\text{m}}{\text{g}}\text{e}^\frac{\text{-gt}}{\text{vm}}\Big)-\frac{\text{V}^2_\text{m}}{\text{g}}$
$=\text{V}_\text{m}\text{t}-\frac{\text{V}^2_\text{m}}{\text{g}}\Big(1-\text{e}^\frac{\text{-gt}}{\text{vm}}\Big)$
$\frac{\text{d}}{\text{dt}}=\text{mgs}=\text{mg}\frac{\text{ds}}{\text{dt}}=\text{mgV}_\text{m}\Big(1-\text{e}^\frac{\text{-gt}}{\text{vm}}\Big)$

$\frac{\text{d}_\text{H}}{\text{dt}}=\text{i}^2\text{R}=\text{R}\Big(\frac{\text{LBV}}{\text{R}}\Big)^2=\frac{\text{L}^2\text{B}^2\text{V}^2}{\text{R}}$
$\Rightarrow\frac{\text{L}^2\text{B}^2}{\text{R}}\text{V}^2_\text{m}\Big(1-\text{e}^\frac{\text{-gt}}{\text{vm}}\Big)^2$
After steady state i.e. $\text{T}\rightarrow\infty$
$\frac{\text{d}}{\text{dt}}\text{mgs}=\text{mgV}_\text{m}$
$\frac{\text{dH}}{\text{dt}}=\frac{\text{L}^2\text{B}^2}{\text{R}}\text{V}^2_\text{m}=\frac{\text{L}^2\text{B}^2}{\text{R}}\text{V}_\text{m}\frac{\text{mgR}}{\text{L}^2\text{B}^2}=\text{mgV}_\text{m}$
Hence after steady state $\frac{\text{dH}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\text{mgs}$

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Question 605 Marks

An LR circuit with emf $\in$ is connected at t = 0.

Find the charge Q which flows through the battery during 0 to t.
Calculate the work done by the battery during this period.
Find the heat developed during this period.
Find the magnetic field energy stored in the circuit at time t.
Verify that the results in the three parts above are consistent with energy conservation.

Answer

Emf = E LR circuit

$\text{dq}=\text{idt}$

$=\text{i}_0\Big(1-\text{e}^\frac{-\text{t}}{\tau}\Big)\text{dt} $

$=\text{i}_0\Big(1-\text{e}^{\text{-IR.L}}\Big)\text{dt} \ \Big[\therefore \ \tau=\frac{\text{L}}{\text{R}}\Big]$

$\text{Q}=\int\limits_\text{0}^\text{t}\text{dq}=\text{i}_0\Bigg[\int\limits_{0}^\text{t}\text{dt}-\int\limits_0^\text{t}\text{e}^\frac{\text{tR}}{\text{L}}\text{dt}\Bigg] $

$=\text{i}_0\Big[\text{t}\Big(\frac{-\text{L}}{\text{R}}\Big)\Big(\text{e}^\frac{-\text{IR}}{\text{L}}\Big)\text{t}_0\Big]$

$=\text{i}_0\Big[\text{t}-\frac{\text{L}}{\text{R}}\Big(1-\text{e}^\frac{-\text{IR}}{\text{L}}\Big)\Big] $

$\text{Q}=\frac{\text{E}}{\text{R}}\Big[\text{t}-\frac{\text{L}}{\text{R}}\Big(1-\text{e}^\frac{-\text{IR}}{\text{L}}\Big)\Big] $

Similarly as we know work done $=\text{Vl}=\text{El}$

$=\text{E}\text{ i}_0 \Big[\text{t}-\frac{\text{L}}{\text{R}}\Big(1-\text{e}^\frac{-\text{lR}}{\text{L}}\Big)\Big]$

$=\frac{\text{E}^2}{\text{R}}\Big[\text{t}-\frac{\text{L}}{\text{R}}\Big(1-\text{e}^\frac{-\text{lR}}{\text{L}}\Big)\Big]$

$\text{H}=\int\limits^\text{t}_0\text{i}^2\text{R}.\text{dt}=\frac{\text{E}^2}{\text{R}^2}.\text{R}.\int\limits^\text{t}_0\Big(1-\text{e}^\frac{-\text{tR}}{\text{L}}\Big)^2.\text{dt}$

$=\frac{\text{E}^2}{\text{R}}\int\limits^\text{t}_0\Big(1+\text{e}^\frac{\big(-2+\text{B}\big)}{\text{L}}-2\text{e}^\frac{-\text{tR}}{\text{L}}\Big).\text{dt}$

$=\frac{\text{E}^2}{\text{R}}\bigg(\text{t}-\frac{\text{L}}{2\text{R}}\text{e}^\frac{-2\text{tR}}{\text{L}}+\frac{2\text{L}}{\text{R}}.\text{e}^\frac{-\text{tR}}{\text{L}}\bigg)^\text{t}_0$

$=\frac{\text{E}^2}{\text{R}}\bigg(\text{t}-\frac{\text{L}}{-2\text{R}}\text{e}^\frac{-2\text{tR}}{\text{L}}+\frac{2\text{L}}{\text{R}}.\text{e}^\frac{-\text{tR}}{\text{L}}\bigg)-\bigg(-\frac{\text{L}}{2\text{R}}+\frac{2\text{L}}{\text{R}}\bigg)$

$=\frac{\text{E}^2}{\text{R}}\bigg[\bigg(\text{t}-\frac{\text{L}}{2\text{R}}\text{x}^2+\frac{2\text{L}}{\text{R}}.\text{x}\bigg)-\frac{3}{2}\frac{\text{L}}{\text{R}}\bigg]$

$=\frac{\text{E}^2}{2}\bigg(\text{t}-\frac{\text{L}}{2\text{R}}\big(\text{x}^2-4\text{x}+3\big)\bigg)$

$\text{E}=\frac{1}{2}\text{Li}^2$

$=\frac{1}{2}\text{L}.\frac{\text{E}^2}{\text{R}^2}.\Big(1-\text{e}^\frac{-\text{tR}}{\text{L}}\Big)^2$ $\bigg[\text{x}=\text{e}^\frac{-\text{tR}}{\text{L}}\bigg]$

$=\frac{\text{LE}^2}{2\text{R}^2}\Big(1-\text{x}\Big)^2$

Total energy used as heat as stored in magnetic field

$=\frac{\text{E}^2}{\text{R}}\text{T}-\frac{\text{E}^2}{\text{R}}.\frac{\text{L}}{2\text{R}}\text{x}^2+\frac{\text{E}^2}{\text{R}}\frac{\text{L}}{\text{r}}.4\text{x}^2$

$-\frac{3\text{L}}{2\text{R}}.\frac{\text{E}^2}{\text{R}}+\frac{\text{LE}^2}{2\text{R}^2}+\frac{\text{LE}^2}{2\text{R}^2}\text{x}^2-\frac{\text{LE}^2}{\text{R}^2}\text{x}$

$=\frac{\text{E}^2}{\text{R}}\text{t}+\frac{\text{E}^2\text{L}}{\text{R}^2}\text{x}-\frac{\text{LE}^2}{\text{R}^2}$

$=\frac{\text{E}^2}{\text{R}}\Big(\text{t}-\frac{\text{L}}{\text{R}}\big(1-\text{x}\big)\Big)$

= Energy drawn from battery

(Hence conservation of energy holds good).

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