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Magnetic field in a plane electromagnetic wave is given by Expression for corresponding electric field will be : Where c is speed of light.
Option: 1
Option: 2
Option: 3
Option: 4

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Posted by

vishal kumar

Two reactions R₁ and R₂ have identical pre-exponential factors. Activation energy of R₁exceeds that of R₂ by 10 kJ mol⁻¹. If k₁ and k₂ are rate constants for reactions R₁ and R₂ respectively at 300 K, then ln(k2/k1) is equal to : (R=8.314 J mol⁻¹K⁻¹)
Option: 1 6
Option: 2 4
Option: 3 8
Option: 4 12

$K=Ae^{\frac{-Ea}{RT}}$

$\frac{K_{1}}{K_{2}}=e\left ( \frac{Ea_{1}-Ea_{2}}{RT} \right )$

Taking log both the sides

$ln\left ( \frac{K_{2}}{K_{1}} \right )=\frac{E_{a1}-E_{a2}}{RT}= \frac{10000}{8.314\times 300}=4$

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Posted by

vishal kumar

A proton with kinetic energy of $1MeV$ moves from south to north. It gets acceleration of $10^{12}m/s^2$ by an applied magnetic field (west to east). The value of magnetic field ( in mT): (Rest mass of a proton is $1.6\times 10^{-27}kg$ )
Option: 1 0.71
Option: 2 71
Option: 3 0.071
Option: 4 7.1

$\begin{array}{l}{\because \mathrm{K.E.}=1.6 \times 10^{-13}=\frac{1}{2} \times 1.6 \times 10^{-27} \mathrm{V}^{2}} \\ \\ {\mathrm{V}=\sqrt{2} \times 10^{7}} \\ \\ {\therefore \mathrm{Bqv}=\mathrm{ma}} \\ \\ {\mathrm{B}=\frac{1.6 \times 10^{-27} \times 10^{12}}{1.6 \times 10^{-19} \times \sqrt{2} \times 10^{7}}} \\ \\ {=0.71 \times 10^{-3} \mathrm{T}} \\ \\ {\text { so } 0.71 \mathrm{mT}}\end{array}$

So option (1) is correct

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Posted by

vishal kumar

A small circular loop of conducting wire has radius a and carries current I. It is placed in a uniform magnetic field B perpendicular to its plane such that when it rotated slightly about its diameter and released, it starts performing simple harmonic motion of time period T. If the mass of the loop is m then :
Option: 1 $T=\sqrt{\frac{\pi m}{IB}}$
Option: 2 $T=\sqrt{\frac{\pi m}{2IB}}$
Option: 3 $T=\sqrt{\frac{2m}{IB}}$
Option: 4 $T=\sqrt{\frac{2\pi m}{IB}}$

$\tau =MB\sin \theta =I\alpha (using \; M=IA)$

$\Rightarrow \pi R^{2}IB\theta =\frac{mR^{2}}{2}\alpha \; (Using \ \alpha =-\omega ^{2}\theta )$

$\Rightarrow \omega =\sqrt{\frac{2\pi IB}{m}}=\frac{2\pi }{t}$

$\Rightarrow T =\sqrt{\frac{2\pi m}{IB}}$

Hence the correct option is (4).

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Posted by

avinash.dongre

An electron gun is placed inside a long solenoid of radius R on its axis. The solenoid has n turns / length and carries a current I. The electron gun shoots an electron along the radius of the solenoid with speed $\nu$ . If the electron does not hit the surface of the solenoid, maximum possible value of $\nu$ is (all symbols have their standard meaning) :
Option: 1 $\frac{2e\mu _{0}nIR}{m}$

Option: 2 $\frac{e\mu _{0}nIR}{2m}$

Option: 3 $\frac{e\mu _{0}nIR}{4m}$

Option: 4 $\frac{e\mu _{0}nIR}{m}$

Malus's Law -

Malus's Law-

This law states that the intensity of the polarized light transmitted through the analyzer varies as the square of the cosine of the angle between the plane of transmission of the analyzer and the plane of the polarizer.

As,

$I=I_{0} \cos ^{2} \theta$

output intensity is given by $I=I_0Cos^2(\theta )$

Initial output intensity=10% of I_0

I.e $\frac{10I_0}{100}=I_0Cos^2(\theta )\Rightarrow \theta =71.57$

Final output intensity=O

means new angle is $90^0$

the angle by which the analyser needs to be rotated further is $90^0-\theta =18.4^0$

So Option (3) is correct.

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Posted by

Ritika Jonwal

A very long wire ABDMNDC is shown in figure carrying current I. AB and BC parts are straight , long and at right angle. At D wire forms circular turn DMND of radius R . AB, BC parts are tangential to circular turn at N and D . Magnetic field at the centre of circle is :
Option: 1 $\frac{\mu _{0}I}{2\pi R}\left ( \pi + \frac{1}{\sqrt{2}}\right )$

Option: 2 $\frac{\mu _{0}I}{2 R}$

Option: 3 $\frac{\mu _{0}I}{2\pi R}\left ( \pi + 1 \right )$

Option: 4 $\frac{\mu _{0}I}{2\pi R}\left ( \pi - \frac{1}{\sqrt{2}}\right )$

As

Magnetic Field due to current in straight wire -

$B=\frac{\mu_{0}}{4 \pi} \cdot \frac{i}{r}\left(\sin \phi_{1}+\sin \phi_{2}\right)$

So

Magnitude of net magnetic field at O is

$\\B_O=B_1+B_2+B_3\\=-\frac{\mu_0i}{4\pi R}(sin90-sin45)+\frac{\mu_0i}{4\pi R}(sin90+sin45)+\frac{\mu_0i}{2 R}\\=\frac{\mu_0i}{4\pi R}(\sqrt{2}+2\pi)$

The direction of magnetic field due to first conductor ( in to the plane) is opposite to that of other two conductors (out of the plane).

Hence the correct option is (1)

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Posted by

vishal kumar

A long, straight wire of radius 'a' carries a current distributed uniformly over its cross-section. The ratio of the magnetic fields due to the wire at distance $\frac{a}{3}$ and 2a, respectively from the axis of the wire is:
Option: 1 $\frac{3}{2}$
Option: 2 2
Option: 3 $\frac{2}{3}$
Option: 4 $\frac{1}{2}$

$B_A = \frac{\mu_0ir}{2\pi a^2} = \frac{\mu_0 i a}{\pi a^2 6} = \frac{\mu_0i}{6\pi a}$

$B_B = \frac{\mu_0i}{2\pi (2a)}$

$\frac{B_A}{B_B} = \frac{4}{6} = \frac{2}{3}$

Hence the option correct option is (3).

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Posted by

avinash.dongre

The electric fields of two plane electromagnetic plane waves in vacuum are given by $\vec{E_1} = E_0\hat{j}\cos(\omega t - kx)$ and $\vec{E_2} = E_0\hat{k}\cos(\omega t - ky)$ At t = 0, a particle of charge q is at origin with velocity $\vec{v} = 0.8c\hat{j}$ (c is the speed of light in vacuum ). The instantaneous force experienced by the particle is:
Option: 1 $E_0 q\left(0.8\hat{i} - \hat{j} + 0.4\hat{k} \right )$
Option: 2 $E_0 q\left(0.4\hat{i} - 3 \hat{j} + 0.8\hat{k} \right )$
Option: 3 $E_0 q\left(-0.8\hat{i} + \hat{j} + \hat{k} \right )$
Option: 4 $E_0 q\left(0.8\hat{i} + \hat{j} + 0.2\hat{k} \right )$

Magnetic field vectors associated with this electromagnetic wave are given by

$\overrightarrow{\mathrm{B}}_{1}=\frac{E_{0}}{\mathrm{c}} \hat{\mathrm{k}} \cos (\mathrm{k} \mathrm{x}-\omega \mathrm{t}) \ \ \& \ \ \overrightarrow{\mathrm{B}}_{2}=\frac{\mathrm{E}_{0}}{\mathrm{c}} \hat{\mathrm{i}} \cos (\mathrm{ky}-\omega \mathrm{t})$