A wave is propagating in a medium of electric dielectric constant 2 and relative magnetic permeability 50. The wave impedance of such a medium is

5 Ω
376.6 Ω
1883 Ω
3776 Ω

If ε₀ and μ₀ are the electric permittivity and magnetic permeability in vacuum, e and mare corresponding quantities in medium, then refractive index of the medium is

\(\sqrt{\frac{\varepsilon }{\varepsilon _{0}}}\)
\(\sqrt{\frac{\varepsilon_{0} \mu }{\varepsilon \mu _{0}}}\)
\(\sqrt{\frac{\varepsilon_{0} \mu_{0} }{\varepsilon \mu }}\)
\(\sqrt{\frac{\varepsilon \mu }{\varepsilon_{0} \mu_{0}}}\)

Solution

The frequency of electromagnetic wave, which best suited to observe a particle of radii 3 × 10^–4cm is of the order of

10¹⁵
10¹⁴
10¹³
10¹²

Solution

Size of particle = 2 ×(3 × 10^–4) = 6 × 10^–4m. To observe a particle, the wavelength of electromagnetic waves must be of the size of particle.

An electromagnetic wave is propagating along Y-axis.Then

oscillating electric field is along X-axis and oscillating magnetic field is along Y-axis
oscillating electric field is along Z-axis and oscillating magnetic field is along X-axis
both oscillating electric and magnetic fields are along Y-axis, but phase difference between them is 90°
both oscillating electric and magnetic fields are mutually perpendicular in arbitrary direction

The energy of electromagnetic wave in vacuum is given by the relation

\(\frac{E^{2}}{2\varepsilon _{0}}+\frac{B^{2}}{2\mu _{0}}\)
\(\frac{1}{2}\varepsilon _{0}E^{2}+\frac{1}{2}\mu _{0}B^{2}\)
\(\frac{E^{2}+B^{2}}{c}\)
\(\frac{1}{2}\varepsilon _{0}E^{2}+\frac{B^{2}}{2\mu _{0}}\)

Solution

The ratio of electric field vector E and magnetic field vector H i.e., (^E⁄_H) has the dimensions of

resistance
inductance
capacitance
product of inductance and capacitance

In a plane electromagnetic wave propagating in space has an electric field of amplitude 9 × 10³ V/m, then the amplitude of the magnetic field is

2.7 × 10¹² T
9.0 × 10^–3 T
3.0 × 10^–4 T
3.0 × 10^–5 T

Solution

A TV tower has a height of 100m. How much population is covered by the TV broadcast if the average population density around the tower is 100 km^–2(radius of the earth= 6.37 × 10⁶ m)

4 lakh
4 billion
40,000
40 lakh

Solution

In an apparatus, the electric field was found to oscillate with an amplitude of 18V/m. The magnitude of the oscillating magnetic field will be

4 × 10^–6 T
6 × 10^–8 T
9 × 10^–9 T
11 × 10^–11 T

Solution

If a source is transmitting electromagnetic wave of frequency 8.2 × 10⁶ Hz, then wavelength of the electromagnetic waves transmitted from the source will be

36.6 m
40.5 m
42.3 m
50.9 m

Solution

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