1
JEE Main 2020 (Online) 9th January Evening Slot
+4
-1
A plane electromagnetic wave is propagating along the direction $${{\widehat i + \widehat j} \over {\sqrt 2 }}$$ , with its polarization along the direction $$\widehat k$$ . The correct form of the magnetic field of the wave would be (here B0 is an appropriate constant) :
A
$${B_0}{{\widehat i - \widehat j} \over {\sqrt 2 }}\cos \left( {\omega t - k{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right)$$
B
$${B_0}{{\widehat i + \widehat j} \over {\sqrt 2 }}\cos \left( {\omega t - k{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right)$$
C
$${B_0}{{\widehat j - \widehat i} \over {\sqrt 2 }}\cos \left( {\omega t + k{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right)$$
D
$${B_0}\widehat k\cos \left( {\omega t - k{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right)$$
2
JEE Main 2020 (Online) 9th January Morning Slot
+4
-1
The electric fields of two plane electromagnetic plane waves in vacuum are given by
$$\overrightarrow {{E_1}} = {E_0}\widehat j\cos \left( {\omega t - kx} \right)$$ and
$$\overrightarrow {{E_2}} = {E_0}\widehat k\cos \left( {\omega t - ky} \right)$$
At t = 0, a particle of charge q is at origin with
a velocity $$\overrightarrow v = 0.8c\widehat j$$ (c is the speed of light in vacuum). The instantaneous force experienced by the particle is :
A
$${E_0}q\left( {0.8\widehat i - \widehat j + 0.4\widehat k} \right)$$
B
$${E_0}q\left( { - 0.8\widehat i + \widehat j + \widehat k} \right)$$
C
$${E_0}q\left( {0.8\widehat i + \widehat j + 0.2\widehat k} \right)$$
D
$${E_0}q\left( {0.4\widehat i - 3\widehat j + 0.8\widehat k} \right)$$
3
JEE Main 2020 (Online) 8th January Evening Slot
+4
-1
A plane electromagnetic wave of frequency 25 GHz is propagating in vacuum along the z-direction. At a particular point in space and time, the magnetic field is given by $$\overrightarrow B = 5 \times {10^{ - 8}}\widehat jT$$. The corresponding electric field $$\overrightarrow E$$ is (speed of light c = 3 × 108 ms–1)
A
15 $$\widehat i$$V / m
B
-15 $$\widehat i$$V / m
C
1.66 × 10–16 $$\widehat i$$V / m
D
-1.66 × 10–16 $$\widehat i$$V / m
4
JEE Main 2020 (Online) 7th January Evening Slot
+4
-1
The electric field of a plane electromagnetic wave is given by
$$\overrightarrow E = {E_0}{{\widehat i + \widehat j} \over {\sqrt 2 }}\cos \left( {kz + \omega t} \right)$$

At t = 0, a positively charged particle is at the point (x, y, z) = $$\left( {0,0,{\pi \over k}} \right)$$.
If its instantaneous velocity at (t = 0) is $${v_0}\widehat k$$ , the force acting on it due to the wave is :
A
parallel to $$\widehat k$$
B
parallel to $${{\widehat i + \widehat j} \over {\sqrt 2 }}$$
C
antiparallel to $${{\widehat i + \widehat j} \over {\sqrt 2 }}$$
D
zero
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