JEE Main 2019 (Online) 8th April Evening Slot | Electromagnetic Waves Question 141 | Physics

JEE Main 2019 (Online) 8th April Evening Slot

MCQ (Single Correct Answer)

-1

The magnetic field of an electromagnetic wave is given by :-

$$\mathop B\limits^ \to = 1.6 \times {10^{ - 6}}\cos \left( {2 \times {{10}^7}z + 6 \times {{10}^{15}}t} \right)\left( {2\mathop i\limits^ \wedge + \mathop j\limits^ \wedge } \right){{Wb} \over {{m^2}}}$$

The associated electric field will be :-

$$\mathop E\limits^ \to = 4.8 \times {10^2}\cos \left( {2 \times {{10}^7}z - 6 \times {{10}^{15}}t} \right)\left( -2{\mathop i\limits^ \wedge + \mathop {j}\limits^ \wedge } \right){V \over m}$$

$$\mathop E\limits^ \to = 4.8 \times {10^2}\cos \left( {2 \times {{10}^7}z - 6 \times {{10}^{15}}t} \right)\left( 2{\mathop i\limits^ \wedge + \mathop {j}\limits^ \wedge } \right){V \over m}$$

$$\mathop E\limits^ \to = 4.8 \times {10^2}\cos \left( {2 \times {{10}^7}z + 6 \times {{10}^{15}}t} \right)\left( {\mathop i\limits^ \wedge - \mathop {2j}\limits^ \wedge } \right){V \over m}$$

$$\mathop E\limits^ \to = 4.8 \times {10^2}\cos \left( {2 \times {{10}^7}z + 6 \times {{10}^{15}}t} \right)\left( -{\mathop i\limits^ \wedge + \mathop {2j}\limits^ \wedge } \right){V \over m}$$

JEE Main 2019 (Online) 8th April Morning Slot

MCQ (Single Correct Answer)

-1

A plane electromagnetic wave travels in free space along the x-direction. The electric field component of the wave at a particular point of space and time is E = 6 V m^–1 along y-direction. Its corresponding magnetic field component, B would be :

2 × 10^–8 T along y-direction

6 × 10^–8 T along z-direction

2 × 10^–8 T along z-direction

6 × 10^–8 T along x-direction

JEE Main 2019 (Online) 12th January Evening Slot

MCQ (Single Correct Answer)

-1

The mean intensity of radiation on the surface of the Sun is about 10⁸ W/m² . The rms value of the corresponding magnetic field is closet to :

10² T

10^$$-$$2 T

10^$$-$$4 T

1 T

JEE Main 2019 (Online) 12th January Morning Slot

MCQ (Single Correct Answer)

-1

A light wave is incident normally on a glass slab of refractive index 1.5. If 4 % of light gets reflected and the amplitude of the electric field of the incident light is 30 V/m, then the amplitude of the electric field for the wave propagating in the glass medium will be :

6 V/m

10 V/m

30 V/m

24 V/m

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