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In a fluorescent lamp choke ( a small transformer) 100 V of reverse voltage is produced when the choke current changes uniformly from 0.25 A to 0 in a duration of 0.025 ms. The self-inductance of the choke (in mH) is estimated to be _____.
Option: 1 10
Option: 2 20
Option: 3 30
Option: 4 40

$V = L\frac{\text d i}{\text d t}\Rightarrow 100 = L\frac{0.25}{0.025} \\ \Rightarrow L = 10 \ \text{mH}$

Posted by

avinash.dongre

A Laser light of wavelength 660 nm is used to weld Retina detachment. If a Laser pulse of width 60 ms and power 0.5 kW is used the approximate number of photons in the pulse are : [Take Planck’s constant h=6.62×10⁻³⁴ Js]
Option: 1 10²⁰
Option: 2 10¹⁸
Option: 3 10²²
Option: 4 10¹⁹

As we discussed in

$E= N \frac{hc}{\lambda }$

or $N=\frac{E\lambda }{hC}\:=\: \frac{0.5\times10^{3}\times60\times10^{-3}\times660\times10^{-9}}{6.62\times10^{-34}\times3\times10^{8}}$

$=\: \frac{10^{-8}\times6.60\times10^{2}}{6.62\times10^{-26}}=10^{20}$

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Posted by

vishal kumar

A signal of frequency 20 kHz and peak voltage of 5 Volt is used to modulate a carrier wave of frequency 1.2 MHz and peak voltage 25 Volts. Choose the correct statement.
Option: 1 Modulation index=5, side frequency bands are at 1400 kHz and 1000 kHz
Option: 2 Modulation index=5, side frequency bands are at 21.2 kHz and 18.8 kHz
Option: 3 Modulation index=0.8, side frequency bands are at 1180 kHz and 1220 kHz
Option: 4 Modulation index=0.2, side frequency bands are at 1220 kHz and 1180 kHz

$\\\text{ Modulate}\ Index: \frac{5}{25}=\frac{1}{5}=0.2\\ \text{Side frequency} (1200+20) H z=1220 k H z \ and \ 1200-20=1180 k H z$

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Posted by

vishal kumar

At time $t=0$ magnetic field of $1000$ Gauss is passing perpendicularly through the area defined by the closed loop shown in the figure. If the magnetic field reduces linearly to $500\: Gauss$ , in the next $5s$ , then induced EMF (in $\mu V$ ) in the loop is :
Option: 1 56

Option: 3 48

Option: 5 36

Option: 7 28

Area of the given loop

$\\A=Area\ of\ rectangle-Area\ of \ 2\ triangles\ with\ height\ 2cm \\ A=16\times4-2\times4=56cm^2$

Magnitude of emf

$\\E=A\frac{dB}{dt}=56\times10^{-4}\times(\frac{(1000-500)\times10^{-4}}{5})\\ E=5600\times10^{-8}V=56\mu \ V$

So the correct graph is given in option 1.

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vishal kumar

A loop ABCDEFA of straight edges has six corner points A(0,0,0), B(5,0,0),C(5,5,0), D(0,5,0), E(0,5,5) and F(0,0,5). The magnetic field in this region is $\vec B=(3\hat{i}+4\hat{j})T$ .The quantity of flux through the loop ABCDEFA (in Wb) is:-
Option: 1 175
Option: 2 60
Option: 3 25
Option: 4 170

Magnetic flux - - wherein

Magnetic flux-

The total number of magnetic lines of force passing normally through an area placed in a magnetic field is equal to
the magnetic flux linked with that area.

I.e for the below figure

Net magnetic flux through the surface is given by

$\phi_B = \oint \vec{B}\cdot \vec{dA}= BA\cos \Theta$

where

$\phi_B=$ Magnetic Flux

$B =$ Magnetic field

$\Theta =$ The angle between area vector and magnetic field vector

Magnetic flux is a scalar quantity.
Unit of magnetic flux -

It's S.I. unit is Weber (wb) or $Tesla\times m^2$ and its C.G.S. unit is maxwell(Mx).

and $1 \ w b=1 \ T m^{2}$ and $1 \ Mx = 10^{-8 } \ wb$

The dimension of magnetic flux is $ML^{2}T^{-2}A^{-1}$
if θ = 0 then $\phi = BA$ and Flux will be positive.
If $\theta =\frac{\pi }{2}$ then Flux will be zero (i.e $\phi = 0$ )

$\vec B=(3\hat{i}+4\hat{j})T$

Total area vectot=area of ABCD+area of DEFA= $5^2\hat{k}+5^2\hat{i}=25 (\hat{i}+\hat{k})$

Total Magnetic flux= $\vec{B}.\vec{A}=(3\hat{i}+4\hat{j}).(25 (\hat{i}+\hat{k}))=(75+100)wb=175 wb$

So option 3 will be correct answer.

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Posted by

Ritika Jonwal

If the magnetic field in a plane electromagnetic wave is given by $\vec B=3\times10^{-8} sin (1.6\times10^3x + 48\times10^{10}t)\ \vec{j} \ T$ ; then what will be the expression for the electric field?

Option: 1 $\vec{E}=\left ( 9\sin \left ( 1.6\times 10^{3}x+48\times 10^{10}t \right )\widehat{k}\; V/m \right )$

Option: 2 $\vec{E}=\left ( 60\sin \left ( 1.6\times 10^{3}x+48\times 10^{10}t \right )\widehat{k}\; V/m \right )$

Option: 3 $\vec{E}=\left ( 3\times 10^{-8}\sin \left ( 1.6\times 10^{3}x+48\times 10^{10}t \right )\widehat{i}\; V/m \right )$

Option: 4 $\vec{E}=\left ( 3\times 10^{-8}\sin \left ( 1.6\times 10^{3}x+48\times 10^{10}t \right )\widehat{j}\; V/m \right )$

Nature of Electromagnetic Waves -

It is also seen from Maxwell’s equations that the magnitude of the electric and the magnetic fields in an electromagnetic wave are related as - $B_{0}= \frac{E_o}{c}$

given, $\vec B=3\times10^{-8} sin (1.6\times10^3x + 48\times10^{10}t)T$

$\begin{aligned} \\ \left | \vec E \right |=BC=3\times10^{-8} sin (1.6\times10^3x + 48\times10^{10}t)\times (3\times10^8)\\ =9 sin (1.6\times10^3x + 48\times10^{10}t)T \end{aligned}$

wave is propagating in -x direction, i.e., in - i direction.

the direction of the EMW wave is in the direction of $\vec E\times \vec B$ .

Since B is in j direction and EMW is in -i direction. Therefore E is in (k) direction.

So Option (1) is correct.

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Posted by

Ritika Jonwal

A LCR circuit behaves like a damped harmonic oscillator. Comparing it with a physical spring-mass damped oscillator having dampling constant 'b', the correct equivalence would be:-

Option: 1 $L\leftrightarrow k,C\leftrightarrow b,R\leftrightarrow m$

Option: 2 $L\leftrightarrow m,C\leftrightarrow k,R\leftrightarrow b$

Option: 3 $L\leftrightarrow m,C\leftrightarrow \frac{1}{k},R\leftrightarrow b$

Option: 4 $L\leftrightarrow \frac{1}{b},C\leftrightarrow \frac{1}{m},R\leftrightarrow \frac{1}{k}$

Series LCR circuit -

Series LCR circuit-

The Figure given above shows a circuit containing a capacitor ,resistor and inductor connected in series through an alternating / sinosoidal voltage source.

As they are in series so the same amount of current will flow in all the three circuit components and for the voltage, vector sum of potential drop across each component would be equal to the applied voltage.

Let 'i' be the amount of current in the circuit at any time and V_L,V_C and V_R the potential drop across L,C and R respectively then
$\begin{array}{l}{\mathrm{v}_{\mathrm{R}}=\mathrm{i} \mathrm{R} \rightarrow \text { Voltage is in phase with i }} \\ \\ {\mathrm{v}_{\mathrm{L}=\mathrm{i} \omega \mathrm{L}} \rightarrow \text { voltage is leading i by } 90^{\circ}} \\ \\ {\mathrm{v}_{\mathrm{c}}=\mathrm{i} / \mathrm{\omega} \mathrm{c} \rightarrow \text { voltage is lagging behind i by } 90^{\circ}}\end{array}\varepsilon$

By all these we can draw phasor diagram as shown below -

One thing should be noticed that we have assumed that V_L is greater than V_C which makes i lags behind V. If V_C > V_L then i lead V. So as per our assumption, there resultant will be (V_L -V_C). So, from the above phasor diagram V will represent resultant of vectors V_R and (V_L -V_C). So the equation become -

$\begin{aligned} V &=\sqrt{V_{R}^{2}+\left(V_{L}-V_{C}\right)^{2}} \\ \\ &=i \sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}} \\ \\ &=i \sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}} \\ \\ &=i Z \\ \text { where, } & \\ Z &=\sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}} \end{aligned}$

Here, Z is called Impedence of this circuit.

Now come to the phase angle. The phase angle for this case is given as -

$\tan \varphi=\frac{V_{L}-V_{C}}{V_{R}}=\frac{X_{L}-X_{C}}{R}=\frac{\omega L-\frac{1}{\omega C}}{R}$

Now from the equation of the phase angle three cases will arise. These three cases are -

(i) When, $\omega L \ > \ \frac{1}{\omega C}$

then, tanφ is positive i.e. φ is positive and voltage leads the current i.
(ii) When $\omega L \ < \ \frac{1}{\omega C}$

then, tanφ is negative i.e. φ is negative and voltage lags behind the current i.
(iii) When $\omega L \ = \ \frac{1}{\omega C}$ ,

then tanφ is zero i.e. φ is zero and voltage and current are in phase. This is called electrical resonance.

-

For the spring-mass damping system, the governing equation is given by

$m \frac{dx^2}{dt^2}+b\frac{dx}{dt}+kx=0...............(1)$

For an LCR damped oscillator, the equation is

$L \frac{dQ^2}{dt^2}+R\frac{dQ}{dt}+\frac{Q}{C}=0...............(2)$

Comparing 1 and 2

$\\L\leftrightarrow m\\C\leftrightarrow \frac{1}{k}\\R\leftrightarrow b$

So the answer is option (3)

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Posted by

Ritika Jonwal

A long solenoid of radius R carries a time(t) - dependent current $I(t)=I_ot(1-t)$ . A ring of radius 2R is placed coaxially near its middle. During the time interval $0\leq t\leq 1$ , the induced current (IR) and the induced EMF (VR) in the ring changes as:-
Option: 1 Direction of $I_{R}$ remains unchanged and $V_{R}$ is zero at $t = 0.25$
Option: 2 Direction of $I_{R}$ remains unchanged and $V_{R}$ is zero at $t = 0.5$
Option: 3 At $t = 0.25$ direction of $I_{R}$ reverses and $V_{R}$ is maximum
Option: 4 At $t = 0.5$ direction of $I_{R}$ reverses and $V_{R}$ is zero

Faraday's law of induction -

Faraday’s First Law-

Whenever the number of magnetic lines of force (Magnetic Flux) passing through a circuit changes an emf called induced emf is produced in the circuit. The induced emf persists only as long as there is a change of flux.

Faraday’s Second Law-

The induced emf is given by the rate of change of magnetic flux linked with the circuit.

i.e Rate of change of magnetic Flux= $\varepsilon = \frac{-d\phi }{dt}$

where $d\phi\rightarrow \phi _{2}-\phi _{1}$ = change in flux

So,

$B=\mu _0nI_0t(1-t)$

using $B_0=\mu _0nI_0$

SO $B=B_0t(1-t)$

Now considering solenoid as ideal solenoid extended up to infinite and ring as its centre

$\phi =BA_s=B_0t(1-t) A_s$

$e=-\frac{d \phi}{dt} =-B_0A_s (1-2t)$

Since induced emf is changing so current will also be changing

because $i=\frac{e}{R}$

So, since direction of emf is changing so direction of current is also changing.

And V_R will be zero at t=0.5 sec

So the option (4) is correct.

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Posted by

Ritika Jonwal

A plane electromagnetic wave is propagating along the direction $\frac{\hat{i}+\hat{j}}{\sqrt{2}},$ with its polarization along the direction $\hat{k}.$ The correct form the magnetic field of the wave would be (here $B_{0}$ is an appropriate constant ) :
Option: 1 $B_{0}\frac{\hat{i}-\hat{j}}{\sqrt{2}}\cos \left ( \omega t-k\frac{\hat{i+\hat{j}}}{\sqrt{2}} \right )$
Option: 2 $B_{0}\frac{\hat{j}-\hat{i}}{\sqrt{2}}\cos \left ( \omega t+k\frac{\hat{i+\hat{j}}}{\sqrt{2}} \right )$
Option: 3 $B_{0}\frac{\hat{i}+\hat{j}}{\sqrt{2}}\cos \left ( \omega t-k\frac{\hat{i+\hat{j}}}{\sqrt{2}} \right )$
Option: 4 $B_{0}\; \hat{k}\; \cos \left ( \omega t-k\frac{\hat{i+\hat{j}}}{\sqrt{2}} \right )$

EM wave is in direction - $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$

As we know that the axis of polarisation of the Em wave is same as Electric field direction that is - $\hat{k}$

$\vec{E}\times \vec{B}$ direction of propagation of EM wave = $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$

$\Rightarrow \vec{k}\times \vec{B}=\frac{\hat{i}+\hat{j}}{\sqrt{2}}$

So B is along $\Rightarrow$ $\frac{\hat{i}-\hat{j}}{\sqrt{2}}$

And the equation of the electromagnetic waves will be in terms of the $\cos (\omega t-\vec{K}\cdot \vec{r})$

So by concluding the above result we can deduce that the option (1) is correct.

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Posted by

avinash.dongre

Consider a coil of wire carrying current I, forming a magnetic dipole. The magnetic flux through an infinite plane that contains the circular coil and excluding the circular coil area is given by $\phi$ i. The magnetic flux through the area given by $\phi$ o. Which of the following is correct?

Option: 1 $\phi$ i = - $\phi_o$
Option: 2 $\phi$ i > $\phi_o$
Option: 3 $\phi$ i < $\phi_o$
Option: 4 $\phi$ i = $-\phi_o$

Magnetic flux - - wherein

Magnetic flux-

The total number of magnetic lines of force passing normally through an area placed in a magnetic field is equal to
the magnetic flux linked with that area.

I.e for the below figure

Net magnetic flux through the surface is given by

$\phi_B = \oint \vec{B}\cdot \vec{dA}= BA\cos \Theta$

where

$\phi_B=$ Magnetic Flux

$B =$ Magnetic field

$\Theta =$ The angle between area vector and magnetic field vector

Flux going right comes back to the left (forms closed loop)

$\int B.dA=0$

Flux inside the coil come bach through outside

$\\\phi _{coil}= -\phi _{out\ side}\\\phi_0= - \phi_i$

So option (4) is correct.

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In a fluorescent lamp choke ( a small transformer) 100 V of reverse voltage is produced when the choke current changes uniformly from 0.25 A to 0 in a duration of 0.025 ms. The self-inductance of the choke (in mH) is estimated to be _____. Option: 1 10 Option: 2 20 Option: 3 30 Option: 4 40

Posted by

avinash.dongre

A Laser light of wavelength 660 nm is used to weld Retina detachment. If a Laser pulse of width 60 ms and power 0.5 kW is used the approximate number of photons in the pulse are : [Take Planck’s constant h=6.62×10−34 Js] Option: 1 1020 Option: 2 1018 Option: 3 1022 Option: 4 1019

Posted by

vishal kumar

Crack CUET with india's "Best Teachers"

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vishal kumar

At time magnetic field of Gauss is passing perpendicularly through the area defined by the closed loop shown in the figure. If the magnetic field reduces linearly to , in the next , then induced EMF (in ) in the loop is : Option: 1 56 Option: 3 48 Option: 5 36 Option: 7 28

Posted by

vishal kumar

JEE Main high-scoring chapters and topics

A loop ABCDEFA of straight edges has six corner points A(0,0,0), B(5,0,0),C(5,5,0), D(0,5,0), E(0,5,5) and F(0,0,5). The magnetic field in this region is .The quantity of flux through the loop ABCDEFA (in Wb) is:- Option: 1 175 Option: 2 60 Option: 3 25 Option: 4 170

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In a fluorescent lamp choke ( a small transformer) 100 V of reverse voltage is produced when the choke current changes uniformly from 0.25 A to 0 in a duration of 0.025 ms. The self-inductance of the choke (in mH) is estimated to be _____.
Option: 1 10
Option: 2 20
Option: 3 30
Option: 4 40

A Laser light of wavelength 660 nm is used to weld Retina detachment. If a Laser pulse of width 60 ms and power 0.5 kW is used the approximate number of photons in the pulse are : [Take Planck’s constant h=6.62×10⁻³⁴ Js]
Option: 1 10²⁰
Option: 2 10¹⁸
Option: 3 10²²
Option: 4 10¹⁹

A loop ABCDEFA of straight edges has six corner points A(0,0,0), B(5,0,0),C(5,5,0), D(0,5,0), E(0,5,5) and F(0,0,5). The magnetic field in this region is $\vec B=(3\hat{i}+4\hat{j})T$ .The quantity of flux through the loop ABCDEFA (in Wb) is:-
Option: 1 175
Option: 2 60
Option: 3 25
Option: 4 170