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Electromagnetic Waves question

2025 · 28 Jan · Shift 2 · Q56
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  5. /2025 · 28 Jan · Shift 2 · Q56

Electromagnetic Waves question

2025 · 28 Jan · Shift 2 · Q56

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
The magnetic field of an E.M. wave is given by B⃗=(32i^+12j^)30sin⁡[ω(t−zc)]\vec{B} = \left( \frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j} \right) 30 \sin \left[ \omega \left( t - \frac{z}{c} \right) \right]B=(23​​i^+21​j^​)30sin[ω(t−cz​)] (S.I. Units). The corresponding electric field in S.I. units is:
  1. A
    E→=(12i^+32j^)30csin⁡[ω(t+zc)]\overrightarrow{\mathrm{E}}=\left(\frac{1}{2} \hat{i}+\frac{\sqrt{3}}{2} \hat{j}\right) 30 \mathrm{c} \sin \left[\omega\left(\mathrm{t}+\frac{z}{\mathrm{c}}\right)\right]E=(21​i^+23​​j^​)30csin[ω(t+cz​)]
  2. B
    E→=(12i^−32j^)30csin⁡[ω(t−zc)]\overrightarrow{\mathrm{E}}=\left(\frac{1}{2} \hat{i}-\frac{\sqrt{3}}{2} \hat{j}\right) 30 \mathrm{c} \sin \left[\omega\left(\mathrm{t}-\frac{z}{\mathrm{c}}\right)\right]E=(21​i^−23​​j^​)30csin[ω(t−cz​)]
  3. C
    E→=(32i^−12j^)30csin⁡[ω(t+zc)]\overrightarrow{\mathrm{E}}=\left(\frac{\sqrt{3}}{2} \hat{i}-\frac{1}{2} \hat{j}\right) 30 \mathrm{c} \sin \left[\omega\left(\mathrm{t}+\frac{z}{\mathrm{c}}\right)\right]E=(23​​i^−21​j^​)30csin[ω(t+cz​)]
  4. D
    E→=(34i^+14j^)30ccos⁡[ω(t−zc)]\overrightarrow{\mathrm{E}}=\left(\frac{3}{4} \hat{i}+\frac{1}{4} \hat{j}\right) 30 \mathrm{c} \cos \left[\omega\left(\mathrm{t}-\frac{z}{\mathrm{c}}\right)\right]E=(43​i^+41​j^​)30ccos[ω(t−cz​)]
View written solutionFree

Correct answer: B

  1. Given magnetic field

The magnetic field is

B⃗=(32i^+12j^)30sin⁡[ω(t−zc)].\vec B = \left(\frac{\sqrt3}{2}\hat i + \frac12 \hat j\right) 30\sin\left[\omega\left(t-\frac{z}{c}\right)\right].B=(23​​i^+21​j^​)30sin[ω(t−cz​)].

So:

  • The wave varies as sin⁡[ω(t−zc)]\sin\left[\omega\left(t-\frac{z}{c}\right)\right]sin[ω(t−cz​)], hence it is propagating in the +z+z+z direction.
  • The magnetic field direction is
B^=32i^+12j^.\hat B = \frac{\sqrt3}{2}\hat i + \frac12\hat j.B^=23​​i^+21​j^​.
  • Magnitude of magnetic field amplitude is
B0=30.B_0 = 30.B0​=30.
  1. Relation between E⃗\vec EE, B⃗\vec BB and direction of propagation

For an electromagnetic wave,

E⃗⊥B⃗,E⃗×B⃗ gives direction of propagation.\vec E \perp \vec B, \qquad \vec E \times \vec B \text{ gives direction of propagation.}E⊥B,E×B gives direction of propagation.

Since propagation is along +z+z+z, we need

E⃗×B⃗=+k^.\vec E \times \vec B = +\hat k.E×B=+k^.

Also, in free space,

E0=cB0=30c.E_0 = cB_0 = 30c.E0​=cB0​=30c.

And E⃗\vec EE and B⃗\vec BB are in phase, so E⃗\vec EE must also contain

sin⁡[ω(t−zc)].\sin\left[\omega\left(t-\frac{z}{c}\right)\right].sin[ω(t−cz​)].
  1. Find the direction of E⃗\vec EE

Let

E⃗=E0(ai^+bj^)sin⁡[ω(t−zc)].\vec E = E_0(a\hat i + b\hat j)\sin\left[\omega\left(t-\frac{z}{c}\right)\right].E=E0​(ai^+bj^​)sin[ω(t−cz​)].

Since E⃗⊥B⃗\vec E \perp \vec BE⊥B,

(ai^+bj^)⋅(32i^+12j^)=0.(a\hat i + b\hat j)\cdot \left(\frac{\sqrt3}{2}\hat i + \frac12\hat j\right)=0.(ai^+bj^​)⋅(23​​i^+21​j^​)=0.

Thus,

a32+b12=0a\frac{\sqrt3}{2} + b\frac12 = 0a23​​+b21​=0 3a+b=0.\sqrt3 a + b = 0.3​a+b=0.

A unit vector satisfying this is

12i^−32j^.\frac12\hat i - \frac{\sqrt3}{2}\hat j.21​i^−23​​j^​.

Check:

(12)(32)+(−32)(12)=0.\left(\frac12\right)\left(\frac{\sqrt3}{2}\right) + \left(-\frac{\sqrt3}{2}\right)\left(\frac12\right)=0.(21​)(23​​)+(−23​​)(21​)=0.
  1. Check propagation direction using cross product

Now compute

(12i^−32j^)×(32i^+12j^).\left(\frac12\hat i - \frac{\sqrt3}{2}\hat j\right) \times \left(\frac{\sqrt3}{2}\hat i + \frac12\hat j\right).(21​i^−23​​j^​)×(23​​i^+21​j^​).

Using i^×j^=k^\hat i \times \hat j = \hat ki^×j^​=k^ and j^×i^=−k^\hat j \times \hat i = -\hat kj^​×i^=−k^:

=12⋅12(i^×j^)+(−32)32(j^×i^)= \frac12\cdot\frac12 (\hat i\times\hat j) + \left(-\frac{\sqrt3}{2}\right)\frac{\sqrt3}{2}(\hat j\times\hat i)=21​⋅21​(i^×j^​)+(−23​​)23​​(j^​×i^) =14k^+(−34)(−k^)= \frac14\hat k + \left(-\frac34\right)(-\hat k)=41​k^+(−43​)(−k^) =14k^+34k^=k^.= \frac14\hat k + \frac34\hat k = \hat k.=41​k^+43​k^=k^.

So this indeed gives propagation along +z+z+z.

  1. Write the electric field

Therefore,

E⃗=(12i^−32j^)30csin⁡[ω(t−zc)].\vec E = \left(\frac12\hat i - \frac{\sqrt3}{2}\hat j\right)30c\sin\left[\omega\left(t-\frac{z}{c}\right)\right].E=(21​i^−23​​j^​)30csin[ω(t−cz​)].
  1. Match with options

This matches Option B.

Final Answer

E⃗=(12i^−32j^)30csin⁡[ω(t−zc)]\boxed{\vec E=\left(\frac12\hat i-\frac{\sqrt3}{2}\hat j\right)30c\sin\left[\omega\left(t-\frac{z}{c}\right)\right]}E=(21​i^−23​​j^​)30csin[ω(t−cz​)]​

So the correct option is B.

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