1
JEE Main 2019 (Online) 10th January Evening Slot
+4
-1
The electric field of a plane polarized electromagnetic wave in free space at time t = 0 is given by an expression $$\overrightarrow E \left( {x,y} \right) = 10\widehat j\cos \left[ {\left( {6x + 8z} \right)} \right].$$ The magnetic field $$\overrightarrow B$$(x,z, t) is given by $$-$$ (c is the velocity of light)
A
$${1 \over c}\left( {6\hat k + 8\widehat i} \right)\cos \left[ {\left( {6x + 8z - 10ct} \right)} \right]$$
B
$${1 \over c}\left( {6\widehat k - 8\widehat i} \right)\cos \left[ {\left( {6x + 8z - 10ct} \right)} \right]$$
C
$${1 \over c}\left( {6\hat k + 8\widehat i} \right)\cos \left[ {\left( {6x - 8z + 10ct} \right)} \right]$$
D
$${1 \over c}\left( {6\hat k - 8\widehat i} \right)\cos \left[ {\left( {6x - 8z + 10ct} \right)} \right]$$
2
JEE Main 2019 (Online) 10th January Morning Slot
+4
-1
If the magnetic field of a plane electromagnetic wave is given by (the speed of light = 3 × 108 B = 100 × 10–6 sin $$\left[ {2\pi \times 2 \times {{10}^{15}}\left( {t - {x \over c}} \right)} \right]$$ then the maximum electric field associated with it is -
A
4.5 $$\times$$ 104 N/C
B
4 $$\times$$ 104 N/C
C
6 $$\times$$ 104 N/C
D
3 $$\times$$ 104 N/C
3
JEE Main 2019 (Online) 9th January Evening Slot
+4
-1
The energy associated with electric field is (UE) and with magnetic field is (UB) for an electromagnetic wave in free space. Then :
A
$${U_E} = {{{U_B}} \over 2}$$
B
$${U_E} > {U_B}$$
C
$${U_E} < {U_B}$$
D
$${U_E} = {U_B}$$
4
JEE Main 2019 (Online) 9th January Morning Slot
+4
-1
A plane electromagnetic wave of frequency 50 MHz travels in free space along the positive x-direction. At a particular point in space and time, $$\overrightarrow E = 6.3\widehat j\,V/m.$$ The corresponding magnetic field $$\overrightarrow {B,}$$ at that point will be :
A
18.9 $$\times$$ 10$$-$$8 $$\widehat k$$T
B
2.1 $$\times$$ 10$$-$$8 $$\widehat k$$T
C
6.3 $$\times$$ 10$$-$$8 $$\widehat k$$T
D
18.9 $$\times$$ 108 $$\widehat k$$T
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