1
JEE Main 2020 (Online) 3rd September Morning Slot
+4
-1
The magnetic field of a plane electromagnetic wave is
$$\overrightarrow B = 3 \times {10^{ - 8}}\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat i$$ T
where c = 3 $$\times$$ 108 ms–1 is the speed of light. The corresponding electric field is :
A
$$\overrightarrow E = - {10^{ - 6}}\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat k$$ V/m
B
$$\overrightarrow E = - 9\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat k$$ V/m
C
$$\overrightarrow E = 9\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat k$$ V/m
D
$$\overrightarrow E = 3 \times {10^{ - 8}}\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat k$$
2
JEE Main 2020 (Online) 2nd September Evening Slot
+4
-1
In a plane electromagnetic wave, the directions of electric field and magnetic field are represented by $$\widehat k$$ and $$2\widehat i - 2\widehat j$$, respectively. What is the unit vector along direction of propagation of the wave?
A
$${1 \over {\sqrt 5 }}\left( {\widehat i + 2\widehat j} \right)$$
B
$${1 \over {\sqrt 5 }}\left( {2\widehat i + \widehat j} \right)$$
C
$${1 \over {\sqrt 2 }}\left( {\widehat i + \widehat j} \right)$$
D
$${1 \over {\sqrt 2 }}\left( {\widehat j + \widehat k} \right)$$
3
JEE Main 2020 (Online) 2nd September Morning Slot
+4
-1
A plane electromagnetic wave, has
frequency of 2.0 $$\times$$ 1010 Hz and its energy density is 1.02 $$\times$$ 10–8 J/m3 in vacuum. The amplitude of the magnetic field of the wave is close to
( $${1 \over {4\pi {\varepsilon _0}}} = 9 \times {10^9}{{N{m^2}} \over {{C^2}}}$$ and speed of light
= 3 $$\times$$ 108 ms–1)
A
190 nT
B
150 nT
C
160 nT
D
180 nT
4
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
$${B_0}\widehat k\cos \left( {\omega t - k{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right)$$
EXAM MAP
Medical
NEET