1
JEE Main 2019 (Online) 11th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A 27 mW laser beam has a cross-sectional area of 10 mm2. The magnitude of the maximum electric field in this electromagnetic wave is given by :
[Given permittivity of space $$ \in $$0 = 9 $$ \times $$ 10–12 SI units, Speed of light c = 3 $$ \times $$ 108 m/s]
A
2 kV/m
B
1 kV/m
C
1.4 kV/m
D
0.7 kV/m
2
JEE Main 2019 (Online) 11th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
An electromagnetic wave of intensity 50 Wm–2 enters in a medium of refractive index 'n' without any loss. The ratio of the magnitudes of electric, and the ratio of the magnitudes of magnetic fields of the wave before and after entering into the medium are respectively, given by:
A
$$\left( {{1 \over {\sqrt n }},{1 \over {\sqrt n }}} \right)$$
B
$$\left( {\sqrt n ,\sqrt n } \right)$$
C
$$\left( {\sqrt n ,{1 \over {\sqrt n }}} \right)$$
D
$$\left( {{1 \over {\sqrt n }},\sqrt n } \right)$$
3
JEE Main 2019 (Online) 10th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
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
$${1 \over c}\left( {6\hat k - 8\widehat i} \right)\cos \left[ {\left( {6x - 8z + 10ct} \right)} \right]$$
4
JEE Main 2019 (Online) 10th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
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
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