DisplacementCurrent
Electromagnetic Wave

155485 On a particular day, the sun delivers an average power of $\left(\frac{6}{\pi} \times 10^{3}\right) \frac{\mathrm{W}}{\mathrm{m}^{2}}$ to the top of
Earth's atmosphere. Find the amplitude of magnetic field for the electromagnetic waves above atmosphere.
(Take $\mu_{0}=4 \pi \times 10^{-7}$ SI unit)

1 $5 \times 10^{-5} \mathrm{~T}$
2 $4 \times 10^{-6} \mathrm{~T}$
3 $6 \times 10^{-6} \mathrm{~T}$
4 $3 \times 10^{-5} \mathrm{~T}$
Electromagnetic Wave

155487 Light wave traveling in air along $x$-direction is given by $E_{y}=540 \sin \pi \times 10^{4}(x-c t) V^{-1}$. Then, the peak value of magnetic field of wave will be (Given $\mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}$ )

1 $18 \times 10^{-7} \mathrm{~T}$
2 $54 \times 10^{-7} \mathrm{~T}$
3 $54 \times 10^{-8} \mathrm{~T}$
4 $18 \times 10^{-8} \mathrm{~T}$
Electromagnetic Wave

155489 The oscillating magnetic field in a plane of electromagnetic wave is given by $B_{y}=5 \times 10^{-6}$ $\sin 1000 \pi\left(5 x-4 \times 10^{8} t\right) T$. The amplitude of electric field will be:

1 $15 \times 10^{2} \mathrm{Vm}^{-1}$
2 $5 \times 10^{-6} \mathrm{Vm}^{-1}$
3 $16 \times 10^{12} \mathrm{Vm}^{-1}$
4 $4 \times 10^{2} \mathrm{Vm}^{-1}$
Electromagnetic Wave

155490 As shown in the figure, after passing through the medium 1. The speed of light $v_{2}$ in medium 2 will be:
(Given, $\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}^{-1}$ )

1 $1.0 \times 10^{8} \mathrm{~ms}^{-1}$
2 $0.5 \times 10^{8} \mathrm{~ms}^{-1}$
3 $1.5 \times 10^{8} \mathrm{~ms}^{-1}$
4 $3.0 \times 10^{8} \mathrm{~ms}^{-1}$
Electromagnetic Wave

155485 On a particular day, the sun delivers an average power of $\left(\frac{6}{\pi} \times 10^{3}\right) \frac{\mathrm{W}}{\mathrm{m}^{2}}$ to the top of
Earth's atmosphere. Find the amplitude of magnetic field for the electromagnetic waves above atmosphere.
(Take $\mu_{0}=4 \pi \times 10^{-7}$ SI unit)

1 $5 \times 10^{-5} \mathrm{~T}$
2 $4 \times 10^{-6} \mathrm{~T}$
3 $6 \times 10^{-6} \mathrm{~T}$
4 $3 \times 10^{-5} \mathrm{~T}$
Electromagnetic Wave

155487 Light wave traveling in air along $x$-direction is given by $E_{y}=540 \sin \pi \times 10^{4}(x-c t) V^{-1}$. Then, the peak value of magnetic field of wave will be (Given $\mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}$ )

1 $18 \times 10^{-7} \mathrm{~T}$
2 $54 \times 10^{-7} \mathrm{~T}$
3 $54 \times 10^{-8} \mathrm{~T}$
4 $18 \times 10^{-8} \mathrm{~T}$
Electromagnetic Wave

155489 The oscillating magnetic field in a plane of electromagnetic wave is given by $B_{y}=5 \times 10^{-6}$ $\sin 1000 \pi\left(5 x-4 \times 10^{8} t\right) T$. The amplitude of electric field will be:

1 $15 \times 10^{2} \mathrm{Vm}^{-1}$
2 $5 \times 10^{-6} \mathrm{Vm}^{-1}$
3 $16 \times 10^{12} \mathrm{Vm}^{-1}$
4 $4 \times 10^{2} \mathrm{Vm}^{-1}$
Electromagnetic Wave

155490 As shown in the figure, after passing through the medium 1. The speed of light $v_{2}$ in medium 2 will be:
(Given, $\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}^{-1}$ )

1 $1.0 \times 10^{8} \mathrm{~ms}^{-1}$
2 $0.5 \times 10^{8} \mathrm{~ms}^{-1}$
3 $1.5 \times 10^{8} \mathrm{~ms}^{-1}$
4 $3.0 \times 10^{8} \mathrm{~ms}^{-1}$
Electromagnetic Wave

155485 On a particular day, the sun delivers an average power of $\left(\frac{6}{\pi} \times 10^{3}\right) \frac{\mathrm{W}}{\mathrm{m}^{2}}$ to the top of
Earth's atmosphere. Find the amplitude of magnetic field for the electromagnetic waves above atmosphere.
(Take $\mu_{0}=4 \pi \times 10^{-7}$ SI unit)

1 $5 \times 10^{-5} \mathrm{~T}$
2 $4 \times 10^{-6} \mathrm{~T}$
3 $6 \times 10^{-6} \mathrm{~T}$
4 $3 \times 10^{-5} \mathrm{~T}$
Electromagnetic Wave

155487 Light wave traveling in air along $x$-direction is given by $E_{y}=540 \sin \pi \times 10^{4}(x-c t) V^{-1}$. Then, the peak value of magnetic field of wave will be (Given $\mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}$ )

1 $18 \times 10^{-7} \mathrm{~T}$
2 $54 \times 10^{-7} \mathrm{~T}$
3 $54 \times 10^{-8} \mathrm{~T}$
4 $18 \times 10^{-8} \mathrm{~T}$
Electromagnetic Wave

155489 The oscillating magnetic field in a plane of electromagnetic wave is given by $B_{y}=5 \times 10^{-6}$ $\sin 1000 \pi\left(5 x-4 \times 10^{8} t\right) T$. The amplitude of electric field will be:

1 $15 \times 10^{2} \mathrm{Vm}^{-1}$
2 $5 \times 10^{-6} \mathrm{Vm}^{-1}$
3 $16 \times 10^{12} \mathrm{Vm}^{-1}$
4 $4 \times 10^{2} \mathrm{Vm}^{-1}$
Electromagnetic Wave

155490 As shown in the figure, after passing through the medium 1. The speed of light $v_{2}$ in medium 2 will be:
(Given, $\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}^{-1}$ )

1 $1.0 \times 10^{8} \mathrm{~ms}^{-1}$
2 $0.5 \times 10^{8} \mathrm{~ms}^{-1}$
3 $1.5 \times 10^{8} \mathrm{~ms}^{-1}$
4 $3.0 \times 10^{8} \mathrm{~ms}^{-1}$
Electromagnetic Wave

155485 On a particular day, the sun delivers an average power of $\left(\frac{6}{\pi} \times 10^{3}\right) \frac{\mathrm{W}}{\mathrm{m}^{2}}$ to the top of
Earth's atmosphere. Find the amplitude of magnetic field for the electromagnetic waves above atmosphere.
(Take $\mu_{0}=4 \pi \times 10^{-7}$ SI unit)

1 $5 \times 10^{-5} \mathrm{~T}$
2 $4 \times 10^{-6} \mathrm{~T}$
3 $6 \times 10^{-6} \mathrm{~T}$
4 $3 \times 10^{-5} \mathrm{~T}$
Electromagnetic Wave

155487 Light wave traveling in air along $x$-direction is given by $E_{y}=540 \sin \pi \times 10^{4}(x-c t) V^{-1}$. Then, the peak value of magnetic field of wave will be (Given $\mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}$ )

1 $18 \times 10^{-7} \mathrm{~T}$
2 $54 \times 10^{-7} \mathrm{~T}$
3 $54 \times 10^{-8} \mathrm{~T}$
4 $18 \times 10^{-8} \mathrm{~T}$
Electromagnetic Wave

155489 The oscillating magnetic field in a plane of electromagnetic wave is given by $B_{y}=5 \times 10^{-6}$ $\sin 1000 \pi\left(5 x-4 \times 10^{8} t\right) T$. The amplitude of electric field will be:

1 $15 \times 10^{2} \mathrm{Vm}^{-1}$
2 $5 \times 10^{-6} \mathrm{Vm}^{-1}$
3 $16 \times 10^{12} \mathrm{Vm}^{-1}$
4 $4 \times 10^{2} \mathrm{Vm}^{-1}$
Electromagnetic Wave

155490 As shown in the figure, after passing through the medium 1. The speed of light $v_{2}$ in medium 2 will be:
(Given, $\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}^{-1}$ )

1 $1.0 \times 10^{8} \mathrm{~ms}^{-1}$
2 $0.5 \times 10^{8} \mathrm{~ms}^{-1}$
3 $1.5 \times 10^{8} \mathrm{~ms}^{-1}$
4 $3.0 \times 10^{8} \mathrm{~ms}^{-1}$