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CBSE Class 12 Physics 2019 Delhi Set 1 Paper

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Question : 15 of 27
Marks: +1, -0
(a) Identify the part of electromagnetic spectrum used in (i) radar and (ii) eye surgery. Write their frequency range.
(b) Prove that the average energy density of the oscillating electric field is equal to that of the oscillating magnetic field.
Solution:  
(a) Identification
Frequency range
(b) Proof
Microwaves: Frequency range
(∼1010  to  1012Hz)(\sim 10^{10} \; \text{to} \; 10^{12} \mathrm{Hz})
Ultraviolet rays: Frequency range
(∼1015  to  1017Hz)(\sim 10^{15} \; \text{to} \; 10^{17} \mathrm{Hz})
(b) Average energy density of the electric field
  =  12ε0E2\;=\;\frac{1}{2} \varepsilon_0 E^2
  =  12ε0(cB)2\;=\;\frac{1}{2} \varepsilon_0 (c B)^2
  =  12ε0  1μ0ε0B2\;=\;\frac{1}{2} \varepsilon_0 \; \frac{1}{\mu_0 \varepsilon_0} B^2
  =  12  B2μ0\;=\;\frac{1}{2} \; \frac{B^2}{\mu_0}
== Average energy density of the magnetic field.
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