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Electromagnetic Waves question

2025 · 23 Jan · Shift 1 · Q64
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  5. /2025 · 23 Jan · Shift 1 · Q64

Electromagnetic Waves question

2025 · 23 Jan · Shift 1 · Q64

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
The electric field of an electromagnetic wave in free space is E→=57cos⁡[7.5×106t−5×10−3(3x+4y)](4i^−3j^)N/C\overrightarrow{\mathrm{E}}=57 \cos \left[7.5 \times 10^6 \mathrm{t}-5 \times 10^{-3}(3 x+4 y)\right](4 \hat{i}-3 \hat{j}) N / CE=57cos[7.5×106t−5×10−3(3x+4y)](4i^−3j^​)N/C. The associated magnetic field in Tesla is
  1. A
    B→=573×108cos⁡[7.5×106t−5×10−3(3x+4y)](k^)\overrightarrow{\mathrm{B}}=\frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 \mathrm{t}-5 \times 10^{-3}(3 x+4 y)\right](\hat{k})B=3×10857​cos[7.5×106t−5×10−3(3x+4y)](k^)
  2. B
    B→=573×108cos⁡[7.5×106t−5×10−3(3x+4y)](5k^)\overrightarrow{\mathrm{B}}=\frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 \mathrm{t}-5 \times 10^{-3}(3 x+4 y)\right](5 \hat{k})B=3×10857​cos[7.5×106t−5×10−3(3x+4y)](5k^)
  3. C
    B→=−573×108cos⁡[7.5×106t−5×10−3(3x+4y)](k^)\overrightarrow{\mathrm{B}}=-\frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 \mathrm{t}-5 \times 10^{-3}(3 x+4 y)\right](\hat{k})B=−3×10857​cos[7.5×106t−5×10−3(3x+4y)](k^)
  4. D
    B→=−573×108cos⁡[7.5×106t−5×10−3(3x+4y)](5k^)\overrightarrow{\mathrm{B}}=-\frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 \mathrm{t}-5 \times 10^{-3}(3 x+4 y)\right](5 \hat{k})B=−3×10857​cos[7.5×106t−5×10−3(3x+4y)](5k^)
View written solutionFree

Correct answer: D

  1. Given electric field
E⃗=57cos⁡[7.5×106t−5×10−3(3x+4y)](4i^−3j^)  N/C\vec E = 57\cos\left[7.5\times 10^6 t - 5\times 10^{-3}(3x+4y)\right](4\hat i-3\hat j)\; \text{N/C}E=57cos[7.5×106t−5×10−3(3x+4y)](4i^−3j^​)N/C

We need the associated magnetic field B⃗\vec BB in free space.

  1. Identify the direction of propagation

For a plane wave,

cos⁡(ωt−k⃗⋅r⃗)\cos(\omega t - \vec k\cdot \vec r)cos(ωt−k⋅r)

Here,

k⃗⋅r⃗=5×10−3(3x+4y)\vec k\cdot \vec r = 5\times 10^{-3}(3x+4y)k⋅r=5×10−3(3x+4y)

So,

k⃗=5×10−3(3i^+4j^)\vec k = 5\times 10^{-3}(3\hat i+4\hat j)k=5×10−3(3i^+4j^​)

Hence the propagation direction is along

3i^+4j^3\hat i+4\hat j3i^+4j^​

The corresponding unit vector is

n^=3i^+4j^5\hat n = \frac{3\hat i+4\hat j}{5}n^=53i^+4j^​​
  1. Direction of magnetic field

For an electromagnetic wave in free space,

B⃗=1c(n^×E⃗)\vec B = \frac{1}{c}(\hat n \times \vec E)B=c1​(n^×E)

Now,

E⃗∝(4i^−3j^)\vec E \propto (4\hat i-3\hat j)E∝(4i^−3j^​)

So compute:

n^×(4i^−3j^)=(3i^+4j^5)×(4i^−3j^)\hat n \times (4\hat i-3\hat j) = \left(\frac{3\hat i+4\hat j}{5}\right)\times (4\hat i-3\hat j)n^×(4i^−3j^​)=(53i^+4j^​​)×(4i^−3j^​)

Using cross products,

(3i^+4j^)×(4i^−3j^)=3i^×4i^+3i^×(−3j^)+4j^×4i^+4j^×(−3j^)(3\hat i+4\hat j)\times(4\hat i-3\hat j) = 3\hat i\times 4\hat i + 3\hat i\times(-3\hat j) + 4\hat j\times 4\hat i + 4\hat j\times(-3\hat j)(3i^+4j^​)×(4i^−3j^​)=3i^×4i^+3i^×(−3j^​)+4j^​×4i^+4j^​×(−3j^​) =0−9k^−16k^+0=−25k^= 0 - 9\hat k -16\hat k +0 = -25\hat k=0−9k^−16k^+0=−25k^

Therefore,

n^×(4i^−3j^)=−25k^5=−5k^\hat n \times (4\hat i-3\hat j)=\frac{-25\hat k}{5}=-5\hat kn^×(4i^−3j^​)=5−25k^​=−5k^

So the magnetic field is in the direction

−k^-\hat k−k^

with factor 555.

  1. Magnitude relation

In free space,

B0=E0cB_0 = \frac{E_0}{c}B0​=cE0​​

Here the magnitude of the electric field amplitude is

E0=57 ∣4i^−3j^∣=57×5E_0 = 57\, |4\hat i-3\hat j| = 57\times 5E0​=57∣4i^−3j^​∣=57×5

Thus,

B0=57×53×108B_0 = \frac{57\times 5}{3\times 10^8}B0​=3×10857×5​

Hence,

B⃗=−573×108cos⁡[7.5×106t−5×10−3(3x+4y)](5k^)\vec B = -\frac{57}{3\times 10^8}\cos\left[7.5\times 10^6 t-5\times 10^{-3}(3x+4y)\right](5\hat k)B=−3×10857​cos[7.5×106t−5×10−3(3x+4y)](5k^)
  1. Match with options

This is exactly Option D.

  1. Comparison with stored answer

Stored correct answer: D
Derived answer: D
So they agree.

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