1
JEE Main 2019 (Online) 12th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A plane electromagnetic wave having a frequency v = 23.9 GHz propagates along the positive z-direction in free space. The peak value of the Electric Field is 60 V/m. Which among the following is the acceptable magnetic field component in the electromagnetic wave ?
A
$$\overrightarrow B $$ = 2 × 10–7 sin(1.5 × 102 x + 0.5 × 1011t) $$\widehat j$$
B
$$\overrightarrow B $$ = 60 sin(0.5 × 103x + 0.5 × 1011t) $$\widehat k$$
C
$$\overrightarrow B $$ = 2 × 10–7 sin(0.5 × 103 z + 1.5 × 1011t) $$\widehat i$$
D
$$\overrightarrow B $$ = 2 × 10–7 sin(0.5 × 103 z - 1.5 × 1011t) $$\widehat i$$
2
JEE Main 2019 (Online) 12th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
An electromagnetic wave is represented by the electric field $$\overrightarrow E = {E_0}\widehat n\sin \left[ {\omega t + \left( {6y - 8z} \right)} \right]$$ . Taking unit vectors in x, y and z directions to be $$\widehat i,\widehat j,\widehat k$$ , the direction of propagation $$\widehat s$$, is :
A
$$\widehat s = {{3\widehat i - 4\widehat j} \over 5}$$
B
$$\widehat s = {{ - 4\widehat k + 3\widehat j} \over 5}$$
C
$$\widehat s = \left( {{{ - 3\widehat j + 4\widehat k} \over 5}} \right)$$
D
$$\widehat s = {{4\widehat j - 3\widehat k} \over 5}$$
3
JEE Main 2019 (Online) 10th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Light is incident normally on a completely absorbing surface with an energy flux of 25 W cm–2. If the surface has an area of 25 cm2, the momentum transferred to the surface in 40 min time duration will be :
A
6.3 × 10–4 Ns
B
5.0 × 10–3 Ns
C
1.4 × 10–6 Ns
D
3.5 × 10–6 Ns
4
JEE Main 2019 (Online) 10th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The electric field of a plane electromagnetic wave is given by
$$\overrightarrow E = {E_0}\widehat i\cos (kz)cos(\omega t)$$
The corresponding magnetic field $$\overrightarrow B $$ is then given by
A
$$\overrightarrow B = {{{E_0}} \over C}\widehat j\sin (kz)\sin (\omega t)$$
B
$$\overrightarrow B = {{{E_0}} \over C}\widehat j\sin (kz)\cos (\omega t)$$
C
$$\overrightarrow B = {{{E_0}} \over C}\widehat j\cos (kz)\sin (\omega t)$$
D
$$\overrightarrow B = {{{E_0}} \over C}\widehat k\sin (kz)\cos (\omega t)$$
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