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Electromagnetic Waves question

2019 · 9 Apr · Shift 1 · Q50
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Electromagnetic Waves question

2019 · 9 Apr · Shift 1 · Q50

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
The magnetic field of a plane electromagnetic wave is given by : B‾=B0i^[cos⁡(kz−ωt)]+Bij^cos⁡(kz+ωt)\overline B = {B_0}\widehat i\left[ {\cos (kz - \omega t)} \right] + {B_i}\widehat j\cos (kz + \omega t)B=B0​i[cos(kz−ωt)]+Bi​j​cos(kz+ωt) B0 = 3 × 10–5 T and B1 = 2 × 10–6 T. The rms value of the force experienced by a stationary charge Q = 10–4 C at z = 0 is closest to :
  1. A
    0.6 N
  2. B
    0.9 N
  3. C
    3 × 10–2 N
  4. D
    0.1 N
View written solutionFree

Correct answer: A

  1. Given magnetic field

The wave magnetic field is

B⃗=B0i^cos⁡(kz−ωt)+B1j^cos⁡(kz+ωt)\vec B = B_0\hat i\cos(kz-\omega t) + B_1\hat j\cos(kz+\omega t)B=B0​i^cos(kz−ωt)+B1​j^​cos(kz+ωt)

with B0=3×10−5 T,B1=2×10−6 T.B_0=3\times 10^{-5}\,\text{T},\qquad B_1=2\times 10^{-6}\,\text{T}.B0​=3×10−5T,B1​=2×10−6T.

We need the force on a stationary charge Q=10−4 CQ=10^{-4}\,\text{C}Q=10−4C at z=0z=0z=0.

  1. Force on a stationary charge

Lorentz force is

F⃗=q(E⃗+v⃗×B⃗).\vec F = q(\vec E + \vec v\times \vec B).F=q(E+v×B).

For a stationary charge, v⃗=0\vec v=0v=0, so

F⃗=qE⃗.\vec F = q\vec E.F=qE.

Hence we first find the electric field corresponding to the given magnetic field.

  1. Find electric field for each wave component

For a plane electromagnetic wave in vacuum,

E=cB,E=cB,E=cB,

and directions satisfy

E⃗×B⃗=direction of propagation.\vec E\times \vec B = \text{direction of propagation}.E×B=direction of propagation.

(i) First component

B⃗1=B0i^cos⁡(kz−ωt)\vec B_1 = B_0\hat i\cos(kz-\omega t)B1​=B0​i^cos(kz−ωt)

This represents a wave traveling in the +z+z+z direction.

We need E⃗1×i^=k^\vec E_1\times \hat i = \hat kE1​×i^=k^. Since

(−j^)×i^=k^,(-\hat j)\times \hat i = \hat k,(−j^​)×i^=k^,

therefore

E⃗1=−cB0j^cos⁡(kz−ωt).\vec E_1 = -cB_0\hat j\cos(kz-\omega t).E1​=−cB0​j^​cos(kz−ωt).

At z=0z=0z=0,

E⃗1=−cB0j^cos⁡ωt.\vec E_1 = -cB_0\hat j\cos\omega t.E1​=−cB0​j^​cosωt.

(ii) Second component

B⃗2=B1j^cos⁡(kz+ωt)\vec B_2 = B_1\hat j\cos(kz+\omega t)B2​=B1​j^​cos(kz+ωt)

This represents a wave traveling in the −z-z−z direction.

We need E⃗2×j^=−k^\vec E_2\times \hat j = -\hat kE2​×j^​=−k^. Since

(−i^)×j^=−k^,(-\hat i)\times \hat j = -\hat k,(−i^)×j^​=−k^,

therefore

E⃗2=−cB1i^cos⁡(kz+ωt).\vec E_2 = -cB_1\hat i\cos(kz+\omega t).E2​=−cB1​i^cos(kz+ωt).

At z=0z=0z=0,

E⃗2=−cB1i^cos⁡ωt.\vec E_2 = -cB_1\hat i\cos\omega t.E2​=−cB1​i^cosωt.
  1. Net electric field at z=0z=0z=0

Thus,

E⃗=−ccos⁡ωt (B1i^+B0j^).\vec E = -c\cos\omega t\,(B_1\hat i + B_0\hat j).E=−ccosωt(B1​i^+B0​j^​).

Its magnitude is

E=ccos⁡ωtB02+B12.E = c\cos\omega t\sqrt{B_0^2+B_1^2}.E=ccosωtB02​+B12​​.

So the force magnitude is

F=qE=qcB02+B12 ∣cos⁡ωt∣.F = qE = q c\sqrt{B_0^2+B_1^2}\,|\cos\omega t|.F=qE=qcB02​+B12​​∣cosωt∣.

The peak force is

F0=qcB02+B12.F_0 = q c\sqrt{B_0^2+B_1^2}.F0​=qcB02​+B12​​.

Hence rms force is

Frms=F02.F_{\rm rms} = \frac{F_0}{\sqrt2}.Frms​=2​F0​​.
  1. Numerical calculation

First,

B02+B12=(3×10−5)2+(2×10−6)2.\sqrt{B_0^2+B_1^2} = \sqrt{(3\times10^{-5})^2+(2\times10^{-6})^2}.B02​+B12​​=(3×10−5)2+(2×10−6)2​.

Since

(3×10−5)2=9×10−10,(3\times10^{-5})^2=9\times10^{-10},(3×10−5)2=9×10−10, (2×10−6)2=4×10−12,(2\times10^{-6})^2=4\times10^{-12},(2×10−6)2=4×10−12,

so

B02+B12=9.04×10−10≈3.01×10−5 T.\sqrt{B_0^2+B_1^2}=\sqrt{9.04\times10^{-10}}\approx 3.01\times10^{-5}\,\text{T}.B02​+B12​​=9.04×10−10​≈3.01×10−5T.

Now,

F0=qcB02+B12=(10−4)(3×108)(3.01×10−5).F_0 = q c\sqrt{B_0^2+B_1^2} = (10^{-4})(3\times10^8)(3.01\times10^{-5}).F0​=qcB02​+B12​​=(10−4)(3×108)(3.01×10−5). F0≈0.903 N.F_0 \approx 0.903\,\text{N}.F0​≈0.903N.

Therefore,

Frms=0.9032≈0.64 N.F_{\rm rms} = \frac{0.903}{\sqrt2} \approx 0.64\,\text{N}.Frms​=2​0.903​≈0.64N.

This is closest to

0.6 N.\boxed{0.6\,\text{N}}.0.6N​.
  1. Option check
  • A: 0.6 N0.6\,\text{N}0.6N ✅ closest
  • B: 0.9 N0.9\,\text{N}0.9N peak value, not rms
  • C: 3×10−2 N3\times10^{-2}\,\text{N}3×10−2N incorrect
  • D: 0.1 N0.1\,\text{N}0.1N incorrect

So the correct option is

A\boxed{\text{A}}A​
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