JEE Physics Electromagnetic Waves MCQs Set 02

Answer: (d) An accelerating charge

Explanation:
1. Only a time-varying current, that is an accelerating charge, radiates.
2. A uniformly moving or static charge only has steady fields.

Teacher's Note:
1. Accelerated charge means radiation.
2. This is the basis of every antenna.

Answer: (c) 2.2 A

Explanation:
1. \( Z=\sqrt{100^2+100^2}=100\sqrt2\ \Omega \).
2. Peak current \( =220\sqrt2/(100\sqrt2)=2.2 \) A.
3. Displacement current between the plates equals the conduction current in the circuit, so its peak is 2.2 A.

Teacher's Note:
1. Displacement current always equals conduction current in a series circuit.
2. The 50 percent charge detail is a distraction.

Answer: (d) \( \dfrac{10^{-2}}{\pi} \) A/m

Explanation:
1. \( H_0=B_0/\mu_0=E_0/(c\mu_0) \).
2. \( H_0=1.2/(3\times10^{8}\times4\pi\times10^{-7})=1.2/377=3.18\times10^{-3} \) A/m.
3. This equals \( 10^{-2}/\pi \) A/m.

Teacher's Note:
1. H is measured in A/m, B in tesla.
2. c times mu0 is 377 ohm.

Answer: (c) 26.5 W/m\(^2\)

Explanation:
1. \( I=\varepsilon_0cE_{rms}^2 \).
2. \( I=8.85\times10^{-12}\times3\times10^{8}\times10^{4}=26.5 \) W/m\(^2\).

Teacher's Note:
1. Intensity uses the rms square of the field.
2. I = c eps0 E_rms^2.

Answer: (d) \( 7\times10^{-6} \) N

Explanation:
1. \( F=IA/c \) for complete absorption.
2. \( F=1400\times1.5/(3\times10^{8})=7\times10^{-6} \) N.

Teacher's Note:
1. Force = pressure x area.
2. Absorbed: I/c; reflected: 2I/c.

Answer: (a) \( 2.83\times10^{-8} \) T

Explanation:
1. \( E_0=\sqrt2E_{rms}=8.49 \) V/m.
2. \( B_0=E_0/c=8.49/(3\times10^{8})=2.83\times10^{-8} \) T.

Teacher's Note:
1. Convert rms to peak by multiplying by root 2.
2. Then divide by c.

Answer: (a) \( 5\times10^{-8} \) T

Explanation:
1. \( I=P/4\pi r^2=15/(16\pi)=0.298 \) W/m\(^2\).
2. \( B_0=\sqrt{2\mu_0I/c}=\sqrt{2\times4\pi\times10^{-7}\times0.298/(3\times10^{8})}=5\times10^{-8} \) T.

Teacher's Note:
1. A point source spreads power over a sphere.
2. Use I = B0^2 c / (2 mu0).

Answer: (d) \( 15\times10^{-7} \) N/m\(^2\)

Explanation:
1. \( p=I/c=450/(3\times10^{8}) \).
2. \( p=1.5\times10^{-6}=15\times10^{-7} \) N/m\(^2\).

Teacher's Note:
1. Absorbing body: p = I/c.
2. Check the power of ten.

Answer: (a) 0.23 mW/m\(^2\)

Explanation:
1. \( I=P/4\pi r^2=18\times10^{-3}/(4\pi\times6.25) \).
2. \( I=2.29\times10^{-4} \) W/m\(^2\) = 0.23 mW/m\(^2\).

Teacher's Note:
1. Divide by the area of the sphere, 4 pi r^2.
2. Watch milli versus unit prefixes.

Answer: (a) \( 3.8\times10^{-5} \) Pa

Explanation:
1. \( I=P/4\pi r^2=6\times10^{5}/(4\pi\times6.25)=7.64\times10^{3} \) W/m\(^2\).
2. With reflected fraction 0.5, \( p=(1+0.5)I/c=1.5\times7.64\times10^{3}/(3\times10^{8}) \).
3. \( p\approx3.8\times10^{-5} \) Pa.

Teacher's Note:
1. Pressure = (1 + reflected fraction) x I / c.

Answer: (b) The +z direction

Explanation:
1. The wave travels along \( \vec E\times\vec B \).
2. \( \hat j\times\hat k=\hat i \), so \( \vec B \) must point along +z.

Teacher's Note:
1. Use the cross product rule E x B = direction of travel.

Answer: (a) 1.65 W/m\(^2\)

Explanation:
1. \( v=1/\sqrt{\mu\varepsilon}=c/2 \).
2. \( I=\tfrac12\varepsilon v E_0^2=\tfrac12\varepsilon_0(c/2)(50)^2 \).
3. \( I=\tfrac12\times8.85\times10^{-12}\times1.5\times10^{8}\times2500\approx1.66 \) W/m\(^2\).

Teacher's Note:
1. In a medium use epsilon and the wave speed v.
2. The answer is about 1.65 W/m^2.

Answer: (a) \( \vec E=753\cos(\omega t-kz)\,\hat j \)

Explanation:
1. \( E_0=377\times2=754\approx753 \) V/m, in phase with H and travelling along +z, so the argument is \( \omega t-kz \).
2. For propagation along +z we need \( \vec E\times\vec H\parallel\hat k \).
3. Here \( \hat j\times(-\hat i)=+\hat k \), so \( \vec E=+753\cos(\omega t-kz)\hat j \).

Teacher's Note:
1. E and H are in phase, with E0/H0 = 377 ohm.
2. Check the sign with the cross product.

Answer: (d) The electric and magnetic field vectors are parallel to each other and perpendicular to the direction of propagation

Explanation:
1. \( \vec E \) and \( \vec B \) are perpendicular to each other and to the direction of travel.
2. The other three statements are true.

Teacher's Note:
1. Fields oscillate in phase.
2. They are mutually perpendicular, not parallel.

Answer: (c) \( LT^{-1} \)

Explanation:
1. \( (\mu_0\varepsilon_0)^{-1/2}=c \), a speed.
2. Its dimensions are \( LT^{-1} \).

Teacher's Note:
1. This expression is just the speed of light.
2. Recognise combinations of constants.

Answer: (c) Frequency

Explanation:
1. The frequency is fixed by the source.
2. Velocity and wavelength change when the wave enters another medium.

Teacher's Note:
1. Frequency never changes on refraction.
2. Wavelength is lambda = v / f.

Answer: (d) Moving along x with frequency \( 10^{6} \) Hz and wavelength 200 m

Explanation:
1. The form \( \cos(\omega t-kx) \) shows travel along +x.
2. \( f=\omega/2\pi=10^{6} \) Hz.
3. \( \lambda=2\pi/k=2\pi/(\pi\times10^{-2})=200 \) m.

Teacher's Note:
1. Minus sign between omega t and kx means +x motion.
2. Frequency is omega over 2 pi.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Both are true.
2. As the capacitor discharges the electric field between the plates changes, and this changing field gives a displacement current and a magnetic field.

Teacher's Note:
1. Maxwell's addition to Ampere's law.
2. Changing E produces B.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Both are true.
2. Circular motion involves centripetal acceleration, and accelerated charges radiate.

Teacher's Note:
1. Cyclotron radiation arises this way.
2. Any acceleration gives radiation.

Answer: (c) \( I/c \)

Explanation:
1. A fully absorbing surface receives momentum at the rate \( I/c \) per unit area.
2. So \( p=I/c \).

Teacher's Note:
1. Check units: W/m^2 divided by m/s gives N/m^2.
2. Reflecting surface gives twice this.