Practice MCQs for JEE Physics Electromagnetic Waves
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Multiple Choice Questions
(a) \( \vec E=4.8\times10^2\cos(2\times10^7z+6\times10^{15}t)(\hat i-2\hat j)\ \text{V m}^{-1} \)
(b) \( \vec E=4.8\times10^2\cos(2\times10^7z-6\times10^{15}t)(-2\hat j+\hat i)\ \text{V m}^{-1} \)
(c) \( \vec E=4.8\times10^2\cos(2\times10^7z+6\times10^{15}t)(-\hat i+2\hat j)\ \text{V m}^{-1} \)
(d) \( \vec E=4.8\times10^2\cos(2\times10^7z-6\times10^{15}t)(2\hat j+\hat i)\ \text{V m}^{-1} \)
Show Answer & Explanation
Answer: (c) \( \vec E=4.8\times10^2\cos(2\times10^7z+6\times10^{15}t)(-\hat i+2\hat j)\ \text{V m}^{-1} \)
Explanation:
1. The phase \( kz+\omega t \) means the wave travels along \( -z \). The magnitude factor is \( cB_0=3\times10^8\times1.6\times10^{-6}=4.8\times10^2 \).
2. E, B and the direction of travel obey \( \vec E\times\vec B\parallel-\hat k \): \( (-\hat i+2\hat j)\times(2\hat i+\hat j)=-5\hat k \), so \( (-\hat i+2\hat j) \) is the correct direction.
Teacher's Note:
1. Write the standard relation for modulation index, bandwidth or line of sight, then substitute carefully.
2. Check the units (kHz, MHz, GHz) before comparing the options.
(a) \( \dfrac1{\sqrt n},\ \sqrt n \)
(b) \( \sqrt n,\ \dfrac1{\sqrt n} \)
(c) \( \sqrt n,\ \sqrt n \)
(d) \( \dfrac1{\sqrt n},\ \dfrac1{\sqrt n} \)
Show Answer & Explanation
Answer: (b) \( \sqrt n,\ \dfrac1{\sqrt n} \)
Explanation:
1. In the medium \( I=\tfrac12\varepsilon_0nc\,E^2 \) is unchanged, so \( E_{med}=\dfrac{E_{air}}{\sqrt n} \) and the ratio \( \dfrac{E_{before}}{E_{after}}=\sqrt n \).
2. \( B=\dfrac{nE}{c} \) in the medium, so \( B_{med}=\sqrt n\,B_{air} \) and the ratio \( \dfrac{B_{before}}{B_{after}}=\dfrac1{\sqrt n} \).
Teacher's Note:
1. Write the standard relation for modulation index, bandwidth or line of sight, then substitute carefully.
2. Check the units (kHz, MHz, GHz) before comparing the options.
(a) \( U_E=U_B \)
(b) \( U_E\gt U_B \)
(c) \( U_E\lt U_B \)
(d) \( U_E=\dfrac{U_B}{2} \)
Show Answer & Explanation
Answer: (a) \( U_E=U_B \)
Explanation:
1. \( u_E=\tfrac12\varepsilon_0E^2 \) and \( u_B=\dfrac{B^2}{2\mu_0} \) with \( E=cB \) and \( c^2=\dfrac1{\mu_0\varepsilon_0} \).
2. Hence \( u_E=u_B \).
Teacher's Note:
1. Write the standard relation for modulation index, bandwidth or line of sight, then substitute carefully.
2. Check the units (kHz, MHz, GHz) before comparing the options.
(a) 3 V/m
(b) 9 V/m
(c) 12 V/m
(d) 6 V/m
Show Answer & Explanation
Answer: (d) 6 V/m
Explanation:
1. \( E_0=cB_0 \).
2. \( E_0=3\times10^8\times20\times10^{-9}=6 \) V/m.
Teacher's Note:
1. Write the standard relation for modulation index, bandwidth or line of sight, then substitute carefully.
2. Check the units (kHz, MHz, GHz) before comparing the options.
(a) nitrogen
(b) boron
(c) carbon
(d) oxygen
Show Answer & Explanation
Answer: (c) carbon
Explanation:
1. Mass number: \( 22-2\times4=14 \). Charge: \( 10-2\times2=6 \).
2. Z = 6 is carbon (C-14).
Teacher's Note:
1. Write the standard relation for modulation index, bandwidth or line of sight, then substitute carefully.
2. Check the units (kHz, MHz, GHz) before comparing the options.
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Electromagnetic Waves Objective Questions & Solutions for JEE Physics
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FAQs
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