JEE Main 2022 (Online) 25th July Evening Shift | Electromagnetic Waves Question 53 | Physics

JEE Main 2022 (Online) 25th July Evening Shift

MCQ (Single Correct Answer)

-1

Light wave travelling in air along x-direction is given by $${E_y} = 540\sin \pi \times {10^4}(x - ct)\,V{m^{ - 1}}$$. Then, the peak value of magnetic field of wave will be (Given c = 3 $$\times$$ 10⁸ ms^$$-$$1)

18 $$\times$$ 10^$$-$$7 T

54 $$\times$$ 10^$$-$$7 T

54 $$\times$$ 10^$$-$$8 T

18 $$\times$$ 10^$$-$$8 T

JEE Main 2022 (Online) 25th July Morning Shift

MCQ (Single Correct Answer)

-1

The rms value of conduction current in a parallel plate capacitor is $$6.9 \,\mu \mathrm{A}$$. The capacity of this capacitor, if it is connected to $$230 \mathrm{~V}$$ ac supply with an angular frequency of $$600 \,\mathrm{rad} / \mathrm{s}$$, will be :

5 pF

50 pF

100 pF

200 pF

JEE Main 2022 (Online) 30th June Morning Shift

MCQ (Single Correct Answer)

-1

An expression for oscillating electric field in a plane electromagnetic wave is given as E_z = 300 sin(5$$\pi$$ $$\times$$ 10³x $$-$$ 3$$\pi$$ $$\times$$ 10¹¹t) Vm^$$-$$1

Then, the value of magnetic field amplitude will be :

(Given : speed of light in Vacuum c = 3 $$\times$$ 10⁸ ms^$$-$$1)

1 $$\times$$ 10^$$-$$6 T

5 $$\times$$ 10^$$-$$6 T

18 $$\times$$ 10⁹ T

21 $$\times$$ 10⁹ T

JEE Main 2022 (Online) 28th June Evening Shift

MCQ (Single Correct Answer)

-1

An EM wave propagating in x-direction has a wavelength of 8 mm. The electric field vibrating y-direction has maximum magnitude of 60 Vm^$$-$$1. Choose the correct equations for electric and magnetic fields if the EM wave is propagating in vacuum :

$${E_y} = 60\sin \left[ {{\pi \over 4} \times {{10}^3}(x - 3 \times {{10}^8}t)} \right]\widehat j\,\,V{m^{ - 1}}$$

$${B_z} = 2\sin \left[ {{\pi \over 4} \times {{10}^3}(x - 3 \times {{10}^8}t)} \right]\widehat k\,\,T$$

$${E_y} = 60\sin \left[ {{\pi \over 4} \times {{10}^3}(x - 3 \times {{10}^8}t)} \right]\widehat j\,\,V{m^{ - 1}}$$

$${B_z} = 2 \times {10^{ - 7}}\sin \left[ {{\pi \over 4} \times {{10}^3}(x - 3 \times {{10}^8}t)} \right]\widehat k\,\,T$$

$${E_y} = 2 \times {10^{ - 7}}\sin \left[ {{\pi \over 4} \times {{10}^3}(x - 3 \times {{10}^8}t)} \right]\widehat j\,\,V{m^{ - 1}}$$

$${B_z} = 60\sin \left[ {{\pi \over 4} \times {{10}^3}(x - 3 \times {{10}^8}t)} \right]\widehat k\,\,T$$

$${E_y} = 2 \times {10^{ - 7}}\sin \left[ {{\pi \over 4} \times {{10}^4}(x - 4 \times {{10}^8}t)} \right]\widehat j\,\,V{m^{ - 1}}$$

$${B_z} = 60\sin \left[ {{\pi \over 4} \times {{10}^4}(x - 4 \times {{10}^8}t)} \right]\widehat k\,\,T$$

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