JEE Main 2020 (Online) 7th January Morning Slot | Electromagnetic Waves Question 117 | Physics

JEE Main 2020 (Online) 7th January Morning Slot

MCQ (Single Correct Answer)

-1

If the magnetic field in a plane electromagnetic wave is given by
$$\overrightarrow B $$ = 3 $$ \times $$ 10^-8 sin(1.6 $$ \times $$ 10³x + 48 $$ \times $$ 10¹⁰t)$$\widehat j$$ T, then what will be expression for electric field ?

$$\overrightarrow E $$ = (9sin(1.6 $$ \times $$ 10³x + 48 $$ \times $$ 10¹⁰t)$$\widehat k$$ V/m)

$$\overrightarrow E $$ = (60sin(1.6 $$ \times $$ 10³x + 48 $$ \times $$ 10¹⁰t)$$\widehat k$$ V/m)

$$\overrightarrow E $$ = (3 $$ \times $$ 10^-8 sin(1.6 $$ \times $$ 10³x + 48 $$ \times $$ 10¹⁰t)$$\widehat i$$ V/m)

$$\overrightarrow E $$ = (3 $$ \times $$ 10^-8 sin(1.6 $$ \times $$ 10³x + 48 $$ \times $$ 10¹⁰t)$$\widehat j$$ V/m)

JEE Main 2019 (Online) 12th April Evening Slot

MCQ (Single Correct Answer)

-1

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 ?

$$\overrightarrow B $$ = 2 × 10^–7 sin(1.5 × 10² x + 0.5 × 10¹¹t) $$\widehat j$$

$$\overrightarrow B $$ = 60 sin(0.5 × 10³x + 0.5 × 10¹¹t) $$\widehat k$$

$$\overrightarrow B $$ = 2 × 10^–7 sin(0.5 × 10³ z + 1.5 × 10¹¹t) $$\widehat i$$

$$\overrightarrow B $$ = 2 × 10^–7 sin(0.5 × 10³ z - 1.5 × 10¹¹t) $$\widehat i$$

JEE Main 2019 (Online) 12th April Morning Slot

MCQ (Single Correct Answer)

-1

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 :

$$\widehat s = {{3\widehat i - 4\widehat j} \over 5}$$

$$\widehat s = {{ - 4\widehat k + 3\widehat j} \over 5}$$

$$\widehat s = \left( {{{ - 3\widehat j + 4\widehat k} \over 5}} \right)$$

$$\widehat s = {{4\widehat j - 3\widehat k} \over 5}$$

JEE Main 2019 (Online) 10th April Evening Slot

MCQ (Single Correct Answer)

-1

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 :

6.3 × 10^–4 Ns

5.0 × 10^–3 Ns

1.4 × 10^–6 Ns

3.5 × 10^–6 Ns

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